Calculus Examples

$f (x) = x^{2} {(x + 3)}^{5}$

Step 1

Find the $x$ values where the second derivative is equal to $0$ .

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Step 1.1

Find the second derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{2}$ and $g (x) = {(x + 3)}^{5}$ .

$x^{2} \frac{d}{d x} [{(x + 3)}^{5}] + {(x + 3)}^{5} \frac{d}{d x} [x^{2}]$

Step 1.1.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{5}$ and $g (x) = x + 3$ .

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Step 1.1.1.2.1

To apply the Chain Rule, set $u$ as $x + 3$ .

$x^{2} (\frac{d}{d u} [u^{5}] \frac{d}{d x} [x + 3]) + {(x + 3)}^{5} \frac{d}{d x} [x^{2}]$

Step 1.1.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 5$ .

$x^{2} (5 u^{4} \frac{d}{d x} [x + 3]) + {(x + 3)}^{5} \frac{d}{d x} [x^{2}]$

Step 1.1.1.2.3

Replace all occurrences of $u$ with $x + 3$ .

$x^{2} (5 {(x + 3)}^{4} \frac{d}{d x} [x + 3]) + {(x + 3)}^{5} \frac{d}{d x} [x^{2}]$

Step 1.1.1.3

Differentiate.

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Step 1.1.1.3.1

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

$x^{2} (5 {(x + 3)}^{4} (\frac{d}{d x} [x] + \frac{d}{d x} [3])) + {(x + 3)}^{5} \frac{d}{d x} [x^{2}]$

Step 1.1.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$x^{2} (5 {(x + 3)}^{4} (1 + \frac{d}{d x} [3])) + {(x + 3)}^{5} \frac{d}{d x} [x^{2}]$

Step 1.1.1.3.3

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$x^{2} (5 {(x + 3)}^{4} (1 + 0)) + {(x + 3)}^{5} \frac{d}{d x} [x^{2}]$

Step 1.1.1.3.4

Simplify the expression.

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Step 1.1.1.3.4.1

Add $1$ and $0$ .

$x^{2} (5 {(x + 3)}^{4} \cdot 1) + {(x + 3)}^{5} \frac{d}{d x} [x^{2}]$

Step 1.1.1.3.4.2

Multiply $5$ by $1$ .

$x^{2} (5 {(x + 3)}^{4}) + {(x + 3)}^{5} \frac{d}{d x} [x^{2}]$

Step 1.1.1.3.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$x^{2} (5 {(x + 3)}^{4}) + {(x + 3)}^{5} (2 x)$

Step 1.1.1.3.6

Move $2$ to the left of ${(x + 3)}^{5}$ .

$x^{2} (5 {(x + 3)}^{4}) + 2 \cdot {(x + 3)}^{5} x$

Step 1.1.1.4

Simplify.

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Step 1.1.1.4.1

Factor $x {(x + 3)}^{4}$ out of $x^{2} (5 {(x + 3)}^{4}) + 2 {(x + 3)}^{5} x$ .

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Step 1.1.1.4.1.1

Factor $x {(x + 3)}^{4}$ out of $x^{2} (5 {(x + 3)}^{4})$ .

$x {(x + 3)}^{4} (x \cdot (5)) + 2 {(x + 3)}^{5} x$

Step 1.1.1.4.1.2

Factor $x {(x + 3)}^{4}$ out of $2 {(x + 3)}^{5} x$ .

$x {(x + 3)}^{4} (x \cdot (5)) + x {(x + 3)}^{4} (2 (x + 3))$

Step 1.1.1.4.1.3

Factor $x {(x + 3)}^{4}$ out of $x {(x + 3)}^{4} (x \cdot (5)) + x {(x + 3)}^{4} (2 (x + 3))$ .

$x {(x + 3)}^{4} (x \cdot (5) + 2 (x + 3))$

Step 1.1.1.4.2

Move $5$ to the left of $x$ .

$f' (x) = x {(x + 3)}^{4} (5 x + 2 (x + 3))$

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

Simplify terms.

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Step 1.1.2.1.1

Simplify each term.

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Step 1.1.2.1.1.1

Apply the distributive property.

$\frac{d}{d x} [x {(x + 3)}^{4} (5 x + 2 x + 2 \cdot 3)]$

Step 1.1.2.1.1.2

Multiply $2$ by $3$ .

$\frac{d}{d x} [x {(x + 3)}^{4} (5 x + 2 x + 6)]$

Step 1.1.2.1.2

Add $5 x$ and $2 x$ .

$\frac{d}{d x} [x {(x + 3)}^{4} (7 x + 6)]$

Step 1.1.2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x {(x + 3)}^{4}$ and $g (x) = 7 x + 6$ .

$x {(x + 3)}^{4} \frac{d}{d x} [7 x + 6] + (7 x + 6) \frac{d}{d x} [x {(x + 3)}^{4}]$

Step 1.1.2.3

Differentiate.

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Step 1.1.2.3.1

By the Sum Rule, the derivative of $7 x + 6$ with respect to $x$ is $\frac{d}{d x} [7 x] + \frac{d}{d x} [6]$ .

$x {(x + 3)}^{4} (\frac{d}{d x} [7 x] + \frac{d}{d x} [6]) + (7 x + 6) \frac{d}{d x} [x {(x + 3)}^{4}]$

Step 1.1.2.3.2

Since $7$ is constant with respect to $x$ , the derivative of $7 x$ with respect to $x$ is $7 \frac{d}{d x} [x]$ .

$x {(x + 3)}^{4} (7 \frac{d}{d x} [x] + \frac{d}{d x} [6]) + (7 x + 6) \frac{d}{d x} [x {(x + 3)}^{4}]$

Step 1.1.2.3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$x {(x + 3)}^{4} (7 \cdot 1 + \frac{d}{d x} [6]) + (7 x + 6) \frac{d}{d x} [x {(x + 3)}^{4}]$

Step 1.1.2.3.4

Multiply $7$ by $1$ .

$x {(x + 3)}^{4} (7 + \frac{d}{d x} [6]) + (7 x + 6) \frac{d}{d x} [x {(x + 3)}^{4}]$

Step 1.1.2.3.5

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$x {(x + 3)}^{4} (7 + 0) + (7 x + 6) \frac{d}{d x} [x {(x + 3)}^{4}]$

Step 1.1.2.3.6

Simplify the expression.

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Step 1.1.2.3.6.1

Add $7$ and $0$ .

$x {(x + 3)}^{4} \cdot 7 + (7 x + 6) \frac{d}{d x} [x {(x + 3)}^{4}]$

Step 1.1.2.3.6.2

Move $7$ to the left of $x {(x + 3)}^{4}$ .

$7 \cdot (x {(x + 3)}^{4}) + (7 x + 6) \frac{d}{d x} [x {(x + 3)}^{4}]$

Step 1.1.2.4

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = {(x + 3)}^{4}$ .

$7 x {(x + 3)}^{4} + (7 x + 6) (x \frac{d}{d x} [{(x + 3)}^{4}] + {(x + 3)}^{4} \frac{d}{d x} [x])$

Step 1.1.2.5

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = x + 3$ .

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Step 1.1.2.5.1

To apply the Chain Rule, set $u$ as $x + 3$ .

$7 x {(x + 3)}^{4} + (7 x + 6) (x (\frac{d}{d u} [u^{4}] \frac{d}{d x} [x + 3]) + {(x + 3)}^{4} \frac{d}{d x} [x])$

Step 1.1.2.5.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 4$ .

$7 x {(x + 3)}^{4} + (7 x + 6) (x (4 u^{3} \frac{d}{d x} [x + 3]) + {(x + 3)}^{4} \frac{d}{d x} [x])$

Step 1.1.2.5.3

Replace all occurrences of $u$ with $x + 3$ .

$7 x {(x + 3)}^{4} + (7 x + 6) (x (4 {(x + 3)}^{3} \frac{d}{d x} [x + 3]) + {(x + 3)}^{4} \frac{d}{d x} [x])$

Step 1.1.2.6

Differentiate.

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Step 1.1.2.6.1

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

$7 x {(x + 3)}^{4} + (7 x + 6) (x (4 {(x + 3)}^{3} (\frac{d}{d x} [x] + \frac{d}{d x} [3])) + {(x + 3)}^{4} \frac{d}{d x} [x])$

Step 1.1.2.6.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$7 x {(x + 3)}^{4} + (7 x + 6) (x (4 {(x + 3)}^{3} (1 + \frac{d}{d x} [3])) + {(x + 3)}^{4} \frac{d}{d x} [x])$

Step 1.1.2.6.3

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$7 x {(x + 3)}^{4} + (7 x + 6) (x (4 {(x + 3)}^{3} (1 + 0)) + {(x + 3)}^{4} \frac{d}{d x} [x])$

Step 1.1.2.6.4

Simplify the expression.

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Step 1.1.2.6.4.1

Add $1$ and $0$ .

$7 x {(x + 3)}^{4} + (7 x + 6) (x (4 {(x + 3)}^{3} \cdot 1) + {(x + 3)}^{4} \frac{d}{d x} [x])$

Step 1.1.2.6.4.2

Multiply $4$ by $1$ .

$7 x {(x + 3)}^{4} + (7 x + 6) (x (4 {(x + 3)}^{3}) + {(x + 3)}^{4} \frac{d}{d x} [x])$

Step 1.1.2.6.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$7 x {(x + 3)}^{4} + (7 x + 6) (x (4 {(x + 3)}^{3}) + {(x + 3)}^{4} \cdot 1)$

Step 1.1.2.6.6

Multiply ${(x + 3)}^{4}$ by $1$ .

$f'' (x) = 7 x {(x + 3)}^{4} + (7 x + 6) (x (4 {(x + 3)}^{3}) + {(x + 3)}^{4})$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $7 x {(x + 3)}^{4} + (7 x + 6) (x (4 {(x + 3)}^{3}) + {(x + 3)}^{4})$ .

$7 x {(x + 3)}^{4} + (7 x + 6) (x (4 {(x + 3)}^{3}) + {(x + 3)}^{4})$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $7 x {(x + 3)}^{4} + (7 x + 6) (x (4 {(x + 3)}^{3}) + {(x + 3)}^{4}) = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$7 x {(x + 3)}^{4} + (7 x + 6) (x (4 {(x + 3)}^{3}) + {(x + 3)}^{4}) = 0$

Step 1.2.2

Simplify $7 x {(x + 3)}^{4} + (7 x + 6) (x (4 {(x + 3)}^{3}) + {(x + 3)}^{4})$ .

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Step 1.2.2.1

Simplify each term.

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Step 1.2.2.1.1

Rewrite using the commutative property of multiplication.

$7 x {(x + 3)}^{4} + (7 x + 6) (4 x {(x + 3)}^{3} + {(x + 3)}^{4}) = 0$

Step 1.2.2.1.2

Expand $(7 x + 6) (4 x {(x + 3)}^{3} + {(x + 3)}^{4})$ using the FOIL Method.

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Step 1.2.2.1.2.1

Apply the distributive property.

$7 x {(x + 3)}^{4} + 7 x (4 x {(x + 3)}^{3} + {(x + 3)}^{4}) + 6 (4 x {(x + 3)}^{3} + {(x + 3)}^{4}) = 0$

Step 1.2.2.1.2.2

Apply the distributive property.

$7 x {(x + 3)}^{4} + 7 x (4 x {(x + 3)}^{3}) + 7 x {(x + 3)}^{4} + 6 (4 x {(x + 3)}^{3} + {(x + 3)}^{4}) = 0$

Step 1.2.2.1.2.3

Apply the distributive property.

$7 x {(x + 3)}^{4} + 7 x (4 x {(x + 3)}^{3}) + 7 x {(x + 3)}^{4} + 6 (4 x {(x + 3)}^{3}) + 6 {(x + 3)}^{4} = 0$

Step 1.2.2.1.3

Simplify each term.

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Step 1.2.2.1.3.1

Multiply $x$ by $x$ by adding the exponents.

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Step 1.2.2.1.3.1.1

Move $x$ .

$7 x {(x + 3)}^{4} + 7 (x \cdot x) (4 {(x + 3)}^{3}) + 7 x {(x + 3)}^{4} + 6 (4 x {(x + 3)}^{3}) + 6 {(x + 3)}^{4} = 0$

Step 1.2.2.1.3.1.2

Multiply $x$ by $x$ .

$7 x {(x + 3)}^{4} + 7 x^{2} (4 {(x + 3)}^{3}) + 7 x {(x + 3)}^{4} + 6 (4 x {(x + 3)}^{3}) + 6 {(x + 3)}^{4} = 0$

Step 1.2.2.1.3.2

Rewrite using the commutative property of multiplication.

$7 x {(x + 3)}^{4} + 7 \cdot 4 x^{2} {(x + 3)}^{3} + 7 x {(x + 3)}^{4} + 6 (4 x {(x + 3)}^{3}) + 6 {(x + 3)}^{4} = 0$

Step 1.2.2.1.3.3

Multiply $7$ by $4$ .

$7 x {(x + 3)}^{4} + 28 x^{2} {(x + 3)}^{3} + 7 x {(x + 3)}^{4} + 6 (4 x {(x + 3)}^{3}) + 6 {(x + 3)}^{4} = 0$

Step 1.2.2.1.3.4

Multiply $4$ by $6$ .

$7 x {(x + 3)}^{4} + 28 x^{2} {(x + 3)}^{3} + 7 x {(x + 3)}^{4} + 24 x {(x + 3)}^{3} + 6 {(x + 3)}^{4} = 0$

Step 1.2.2.2

Add $7 x {(x + 3)}^{4}$ and $7 x {(x + 3)}^{4}$ .

$14 x {(x + 3)}^{4} + 28 x^{2} {(x + 3)}^{3} + 24 x {(x + 3)}^{3} + 6 {(x + 3)}^{4} = 0$

Step 1.2.3

Factor the left side of the equation.

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Step 1.2.3.1

Factor $2 {(x + 3)}^{3}$ out of $14 x {(x + 3)}^{4} + 28 x^{2} {(x + 3)}^{3} + 24 x {(x + 3)}^{3} + 6 {(x + 3)}^{4}$ .

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Step 1.2.3.1.1

Factor $2 {(x + 3)}^{3}$ out of $14 x {(x + 3)}^{4}$ .

$2 {(x + 3)}^{3} (7 x (x + 3)) + 28 x^{2} {(x + 3)}^{3} + 24 x {(x + 3)}^{3} + 6 {(x + 3)}^{4} = 0$

Step 1.2.3.1.2

Factor $2 {(x + 3)}^{3}$ out of $28 x^{2} {(x + 3)}^{3}$ .

$2 {(x + 3)}^{3} (7 x (x + 3)) + 2 {(x + 3)}^{3} (14 x^{2}) + 24 x {(x + 3)}^{3} + 6 {(x + 3)}^{4} = 0$

Step 1.2.3.1.3

Factor $2 {(x + 3)}^{3}$ out of $24 x {(x + 3)}^{3}$ .

$2 {(x + 3)}^{3} (7 x (x + 3)) + 2 {(x + 3)}^{3} (14 x^{2}) + 2 {(x + 3)}^{3} (12 x) + 6 {(x + 3)}^{4} = 0$

Step 1.2.3.1.4

Factor $2 {(x + 3)}^{3}$ out of $6 {(x + 3)}^{4}$ .

$2 {(x + 3)}^{3} (7 x (x + 3)) + 2 {(x + 3)}^{3} (14 x^{2}) + 2 {(x + 3)}^{3} (12 x) + 2 {(x + 3)}^{3} (3 (x + 3)) = 0$

Step 1.2.3.1.5

Factor $2 {(x + 3)}^{3}$ out of $2 {(x + 3)}^{3} (7 x (x + 3)) + 2 {(x + 3)}^{3} (14 x^{2})$ .

$2 {(x + 3)}^{3} (7 x (x + 3) + 14 x^{2}) + 2 {(x + 3)}^{3} (12 x) + 2 {(x + 3)}^{3} (3 (x + 3)) = 0$

Step 1.2.3.1.6

Factor $2 {(x + 3)}^{3}$ out of $2 {(x + 3)}^{3} (7 x (x + 3) + 14 x^{2}) + 2 {(x + 3)}^{3} (12 x)$ .

$2 {(x + 3)}^{3} (7 x (x + 3) + 14 x^{2} + 12 x) + 2 {(x + 3)}^{3} (3 (x + 3)) = 0$

Step 1.2.3.1.7

Factor $2 {(x + 3)}^{3}$ out of $2 {(x + 3)}^{3} (7 x (x + 3) + 14 x^{2} + 12 x) + 2 {(x + 3)}^{3} (3 (x + 3))$ .

$2 {(x + 3)}^{3} (7 x (x + 3) + 14 x^{2} + 12 x + 3 (x + 3)) = 0$

Step 1.2.3.2

Factor $7 x$ out of $7 x (x + 3) + 14 x^{2}$ .

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Step 1.2.3.2.1

Factor $7 x$ out of $14 x^{2}$ .

$2 {(x + 3)}^{3} (7 x (x + 3) + 7 x (2 x) + 12 x + 3 (x + 3)) = 0$

Step 1.2.3.2.2

Factor $7 x$ out of $7 x (x + 3) + 7 x (2 x)$ .

$2 {(x + 3)}^{3} (7 x (x + 3 + 2 x) + 12 x + 3 (x + 3)) = 0$

Step 1.2.3.3

Add $x$ and $2 x$ .

$2 {(x + 3)}^{3} (7 x (3 x + 3) + 12 x + 3 (x + 3)) = 0$

Step 1.2.3.4

Factor.

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Step 1.2.3.4.1

Factor $3$ out of $3 x + 3$ .

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Step 1.2.3.4.1.1

Factor $3$ out of $3 x$ .

$2 {(x + 3)}^{3} (7 x (3 (x) + 3) + 12 x + 3 (x + 3)) = 0$

Step 1.2.3.4.1.2

Factor $3$ out of $3$ .

$2 {(x + 3)}^{3} (7 x (3 (x) + 3 (1)) + 12 x + 3 (x + 3)) = 0$

Step 1.2.3.4.1.3

Factor $3$ out of $3 (x) + 3 (1)$ .

$2 {(x + 3)}^{3} (7 x (3 (x + 1)) + 12 x + 3 (x + 3)) = 0$

Step 1.2.3.4.2

Remove unnecessary parentheses.

$2 {(x + 3)}^{3} (7 x \cdot (3 (x + 1)) + 12 x + 3 (x + 3)) = 0$

Step 1.2.3.5

Multiply $3$ by $7$ .

$2 {(x + 3)}^{3} (21 x (x + 1) + 12 x + 3 (x + 3)) = 0$

Step 1.2.3.6

Factor $3$ out of $12 x + 3 (x + 3)$ .

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Step 1.2.3.6.1

Factor $3$ out of $12 x$ .

$2 {(x + 3)}^{3} (21 x (x + 1) + 3 (4 x) + 3 (x + 3)) = 0$

Step 1.2.3.6.2

Factor $3$ out of $3 (4 x) + 3 (x + 3)$ .

$2 {(x + 3)}^{3} (21 x (x + 1) + 3 (4 x + x + 3)) = 0$

Step 1.2.3.7

Add $4 x$ and $x$ .

$2 {(x + 3)}^{3} (21 x (x + 1) + 3 (5 x + 3)) = 0$

Step 1.2.3.8

Factor $3$ out of $21 x (x + 1) + 3 (5 x + 3)$ .

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Step 1.2.3.8.1

Factor $3$ out of $21 x (x + 1)$ .

$2 {(x + 3)}^{3} (3 (7 x (x + 1)) + 3 (5 x + 3)) = 0$

Step 1.2.3.8.2

Factor $3$ out of $3 (7 x (x + 1)) + 3 (5 x + 3)$ .

$2 {(x + 3)}^{3} (3 (7 x (x + 1) + 5 x + 3)) = 0$

Step 1.2.3.9

Apply the distributive property.

$2 {(x + 3)}^{3} (3 (7 x \cdot x + 7 x \cdot 1 + 5 x + 3)) = 0$

Step 1.2.3.10

Multiply $x$ by $x$ by adding the exponents.

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Step 1.2.3.10.1

Move $x$ .

$2 {(x + 3)}^{3} (3 (7 (x \cdot x) + 7 x \cdot 1 + 5 x + 3)) = 0$

Step 1.2.3.10.2

Multiply $x$ by $x$ .

$2 {(x + 3)}^{3} (3 (7 x^{2} + 7 x \cdot 1 + 5 x + 3)) = 0$

Step 1.2.3.11

Multiply $7$ by $1$ .

$2 {(x + 3)}^{3} (3 (7 x^{2} + 7 x + 5 x + 3)) = 0$

Step 1.2.3.12

Factor.

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Step 1.2.3.12.1

Add $7 x$ and $5 x$ .

$2 {(x + 3)}^{3} (3 (7 x^{2} + 12 x + 3)) = 0$

Step 1.2.3.12.2

Remove unnecessary parentheses.

$2 {(x + 3)}^{3} \cdot (3 (7 x^{2} + 12 x + 3)) = 0$

Step 1.2.3.13

Multiply $3$ by $2$ .

$6 {(x + 3)}^{3} (7 x^{2} + 12 x + 3) = 0$

Step 1.2.4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

${(x + 3)}^{3} = 0$

$7 x^{2} + 12 x + 3 = 0$

Step 1.2.5

Set ${(x + 3)}^{3}$ equal to $0$ and solve for $x$ .

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Step 1.2.5.1

Set ${(x + 3)}^{3}$ equal to $0$ .

${(x + 3)}^{3} = 0$

Step 1.2.5.2

Solve ${(x + 3)}^{3} = 0$ for $x$ .

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Step 1.2.5.2.1

Set the $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 1.2.5.2.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 1.2.6

Set $7 x^{2} + 12 x + 3$ equal to $0$ and solve for $x$ .

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Step 1.2.6.1

Set $7 x^{2} + 12 x + 3$ equal to $0$ .

$7 x^{2} + 12 x + 3 = 0$

Step 1.2.6.2

Solve $7 x^{2} + 12 x + 3 = 0$ for $x$ .

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Step 1.2.6.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 1.2.6.2.2

Substitute the values $a = 7$ , $b = 12$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (7 \cdot 3)}}{2 \cdot 7}$

Step 1.2.6.2.3

Simplify.

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Step 1.2.6.2.3.1

Simplify the numerator.

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Step 1.2.6.2.3.1.1

Raise $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 7 \cdot 3}}{2 \cdot 7}$

Step 1.2.6.2.3.1.2

Multiply $- 4 \cdot 7 \cdot 3$ .

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Step 1.2.6.2.3.1.2.1

Multiply $- 4$ by $7$ .

$x = \frac{- 12 \pm \sqrt{144 - 28 \cdot 3}}{2 \cdot 7}$

Step 1.2.6.2.3.1.2.2

Multiply $- 28$ by $3$ .

$x = \frac{- 12 \pm \sqrt{144 - 84}}{2 \cdot 7}$

Step 1.2.6.2.3.1.3

Subtract $84$ from $144$ .

$x = \frac{- 12 \pm \sqrt{60}}{2 \cdot 7}$

Step 1.2.6.2.3.1.4

Rewrite $60$ as $2^{2} \cdot 15$ .

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Step 1.2.6.2.3.1.4.1

Factor $4$ out of $60$ .

$x = \frac{- 12 \pm \sqrt{4 (15)}}{2 \cdot 7}$

Step 1.2.6.2.3.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 12 \pm \sqrt{2^{2} \cdot 15}}{2 \cdot 7}$

Step 1.2.6.2.3.1.5

Pull terms out from under the radical.

$x = \frac{- 12 \pm 2 \sqrt{15}}{2 \cdot 7}$

Step 1.2.6.2.3.2

Multiply $2$ by $7$ .

$x = \frac{- 12 \pm 2 \sqrt{15}}{14}$

Step 1.2.6.2.3.3

Simplify $\frac{- 12 \pm 2 \sqrt{15}}{14}$ .

$x = \frac{- 6 \pm \sqrt{15}}{7}$

Step 1.2.6.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 1.2.6.2.4.1

Simplify the numerator.

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Step 1.2.6.2.4.1.1

Raise $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 7 \cdot 3}}{2 \cdot 7}$

Step 1.2.6.2.4.1.2

Multiply $- 4 \cdot 7 \cdot 3$ .

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Step 1.2.6.2.4.1.2.1

Multiply $- 4$ by $7$ .

$x = \frac{- 12 \pm \sqrt{144 - 28 \cdot 3}}{2 \cdot 7}$

Step 1.2.6.2.4.1.2.2

Multiply $- 28$ by $3$ .

$x = \frac{- 12 \pm \sqrt{144 - 84}}{2 \cdot 7}$

Step 1.2.6.2.4.1.3

Subtract $84$ from $144$ .

$x = \frac{- 12 \pm \sqrt{60}}{2 \cdot 7}$

Step 1.2.6.2.4.1.4

Rewrite $60$ as $2^{2} \cdot 15$ .

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Step 1.2.6.2.4.1.4.1

Factor $4$ out of $60$ .

$x = \frac{- 12 \pm \sqrt{4 (15)}}{2 \cdot 7}$

Step 1.2.6.2.4.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 12 \pm \sqrt{2^{2} \cdot 15}}{2 \cdot 7}$

Step 1.2.6.2.4.1.5

Pull terms out from under the radical.

$x = \frac{- 12 \pm 2 \sqrt{15}}{2 \cdot 7}$

Step 1.2.6.2.4.2

Multiply $2$ by $7$ .

$x = \frac{- 12 \pm 2 \sqrt{15}}{14}$

Step 1.2.6.2.4.3

Simplify $\frac{- 12 \pm 2 \sqrt{15}}{14}$ .

$x = \frac{- 6 \pm \sqrt{15}}{7}$

Step 1.2.6.2.4.4

Change the $\pm$ to $+$ .

$x = \frac{- 6 + \sqrt{15}}{7}$

Step 1.2.6.2.4.5

Rewrite $- 6$ as $- 1 (6)$ .

$x = \frac{- 1 \cdot 6 + \sqrt{15}}{7}$

Step 1.2.6.2.4.6

Factor $- 1$ out of $\sqrt{15}$ .

$x = \frac{- 1 \cdot 6 - 1 (- \sqrt{15})}{7}$

Step 1.2.6.2.4.7

Factor $- 1$ out of $- 1 (6) - 1 (- \sqrt{15})$ .

$x = \frac{- 1 (6 - \sqrt{15})}{7}$

Step 1.2.6.2.4.8

Move the negative in front of the fraction.

$x = - \frac{6 - \sqrt{15}}{7}$

Step 1.2.6.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 1.2.6.2.5.1

Simplify the numerator.

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Step 1.2.6.2.5.1.1

Raise $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 7 \cdot 3}}{2 \cdot 7}$

Step 1.2.6.2.5.1.2

Multiply $- 4 \cdot 7 \cdot 3$ .

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Step 1.2.6.2.5.1.2.1

Multiply $- 4$ by $7$ .

$x = \frac{- 12 \pm \sqrt{144 - 28 \cdot 3}}{2 \cdot 7}$

Step 1.2.6.2.5.1.2.2

Multiply $- 28$ by $3$ .

$x = \frac{- 12 \pm \sqrt{144 - 84}}{2 \cdot 7}$

Step 1.2.6.2.5.1.3

Subtract $84$ from $144$ .

$x = \frac{- 12 \pm \sqrt{60}}{2 \cdot 7}$

Step 1.2.6.2.5.1.4

Rewrite $60$ as $2^{2} \cdot 15$ .

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Step 1.2.6.2.5.1.4.1

Factor $4$ out of $60$ .

$x = \frac{- 12 \pm \sqrt{4 (15)}}{2 \cdot 7}$

Step 1.2.6.2.5.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 12 \pm \sqrt{2^{2} \cdot 15}}{2 \cdot 7}$

Step 1.2.6.2.5.1.5

Pull terms out from under the radical.

$x = \frac{- 12 \pm 2 \sqrt{15}}{2 \cdot 7}$

Step 1.2.6.2.5.2

Multiply $2$ by $7$ .

$x = \frac{- 12 \pm 2 \sqrt{15}}{14}$

Step 1.2.6.2.5.3

Simplify $\frac{- 12 \pm 2 \sqrt{15}}{14}$ .

$x = \frac{- 6 \pm \sqrt{15}}{7}$

Step 1.2.6.2.5.4

Change the $\pm$ to $-$ .

$x = \frac{- 6 - \sqrt{15}}{7}$

Step 1.2.6.2.5.5

Rewrite $- 6$ as $- 1 (6)$ .

$x = \frac{- 1 \cdot 6 - \sqrt{15}}{7}$

Step 1.2.6.2.5.6

Factor $- 1$ out of $- \sqrt{15}$ .

$x = \frac{- 1 \cdot 6 - (\sqrt{15})}{7}$

Step 1.2.6.2.5.7

Factor $- 1$ out of $- 1 (6) - (\sqrt{15})$ .

$x = \frac{- 1 (6 + \sqrt{15})}{7}$

Step 1.2.6.2.5.8

Move the negative in front of the fraction.

$x = - \frac{6 + \sqrt{15}}{7}$

Step 1.2.6.2.6

The final answer is the combination of both solutions.

$x = - \frac{6 - \sqrt{15}}{7}, - \frac{6 + \sqrt{15}}{7}$

Step 1.2.7

The final solution is all the values that make $6 {(x + 3)}^{3} (7 x^{2} + 12 x + 3) = 0$ true.

$x = - 3, - \frac{6 - \sqrt{15}}{7}, - \frac{6 + \sqrt{15}}{7}$

Step 2

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, - 3) \cup (- 3, - \frac{6 + \sqrt{15}}{7}) \cup (- \frac{6 + \sqrt{15}}{7}, - \frac{6 - \sqrt{15}}{7}) \cup (- \frac{6 - \sqrt{15}}{7}, \infty)$

Step 4

Substitute any number from the interval $(- \infty, - 3)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

Replace the variable $x$ with $- 6$ in the expression.

$f'' (- 6) = 7 (- 6) {((- 6) + 3)}^{4} + (7 (- 6) + 6) ((- 6) (4 {((- 6) + 3)}^{3}) + {((- 6) + 3)}^{4})$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply $7$ by $- 6$ .

$f'' (- 6) = - 42 {((- 6) + 3)}^{4} + (7 (- 6) + 6) ((- 6) (4 {((- 6) + 3)}^{3}) + {((- 6) + 3)}^{4})$

Step 4.2.1.2

Add $- 6$ and $3$ .

$f'' (- 6) = - 42 {(- 3)}^{4} + (7 (- 6) + 6) ((- 6) (4 {((- 6) + 3)}^{3}) + {((- 6) + 3)}^{4})$

Step 4.2.1.3

Raise $- 3$ to the power of $4$ .

$f'' (- 6) = - 42 \cdot 81 + (7 (- 6) + 6) ((- 6) (4 {((- 6) + 3)}^{3}) + {((- 6) + 3)}^{4})$

Step 4.2.1.4

Multiply $- 42$ by $81$ .

$f'' (- 6) = - 3402 + (7 (- 6) + 6) ((- 6) (4 {((- 6) + 3)}^{3}) + {((- 6) + 3)}^{4})$

Step 4.2.1.5

Multiply $7$ by $- 6$ .

$f'' (- 6) = - 3402 + (- 42 + 6) ((- 6) (4 {((- 6) + 3)}^{3}) + {((- 6) + 3)}^{4})$

Step 4.2.1.6

Add $- 42$ and $6$ .

$f'' (- 6) = - 3402 - 36 ((- 6) (4 {((- 6) + 3)}^{3}) + {((- 6) + 3)}^{4})$

Step 4.2.1.7

Simplify each term.

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Step 4.2.1.7.1

Add $- 6$ and $3$ .

$f'' (- 6) = - 3402 - 36 (- 6 (4 {(- 3)}^{3}) + {((- 6) + 3)}^{4})$

Step 4.2.1.7.2

Raise $- 3$ to the power of $3$ .

$f'' (- 6) = - 3402 - 36 (- 6 (4 \cdot - 27) + {((- 6) + 3)}^{4})$

Step 4.2.1.7.3

Multiply $- 6 (4 \cdot - 27)$ .

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Step 4.2.1.7.3.1

Multiply $4$ by $- 27$ .

$f'' (- 6) = - 3402 - 36 (- 6 \cdot - 108 + {((- 6) + 3)}^{4})$

Step 4.2.1.7.3.2

Multiply $- 6$ by $- 108$ .

$f'' (- 6) = - 3402 - 36 (648 + {((- 6) + 3)}^{4})$

Step 4.2.1.7.4

Add $- 6$ and $3$ .

$f'' (- 6) = - 3402 - 36 (648 + {(- 3)}^{4})$

Step 4.2.1.7.5

Raise $- 3$ to the power of $4$ .

$f'' (- 6) = - 3402 - 36 (648 + 81)$

Step 4.2.1.8

Add $648$ and $81$ .

$f'' (- 6) = - 3402 - 36 \cdot 729$

Step 4.2.1.9

Multiply $- 36$ by $729$ .

$f'' (- 6) = - 3402 - 26244$

Step 4.2.2

Subtract $26244$ from $- 3402$ .

$f'' (- 6) = - 29646$

Step 4.2.3

The final answer is $- 29646$ .

$- 29646$

Step 4.3

The graph is concave down on the interval $(- \infty, - 3)$ because $f'' (- 6)$ is negative.

Concave down on $(- \infty, - 3)$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(- 3, - \frac{6 + \sqrt{15}}{7})$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

Replace the variable $x$ with $- 2$ in the expression.

$f'' (- 2) = 7 (- 2) {((- 2) + 3)}^{4} + (7 (- 2) + 6) ((- 2) (4 {((- 2) + 3)}^{3}) + {((- 2) + 3)}^{4})$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Multiply $7$ by $- 2$ .

$f'' (- 2) = - 14 {((- 2) + 3)}^{4} + (7 (- 2) + 6) ((- 2) (4 {((- 2) + 3)}^{3}) + {((- 2) + 3)}^{4})$

Step 5.2.1.2

Add $- 2$ and $3$ .

$f'' (- 2) = - 14 \cdot 1^{4} + (7 (- 2) + 6) ((- 2) (4 {((- 2) + 3)}^{3}) + {((- 2) + 3)}^{4})$

Step 5.2.1.3

One to any power is one.

$f'' (- 2) = - 14 \cdot 1 + (7 (- 2) + 6) ((- 2) (4 {((- 2) + 3)}^{3}) + {((- 2) + 3)}^{4})$

Step 5.2.1.4

Multiply $- 14$ by $1$ .

$f'' (- 2) = - 14 + (7 (- 2) + 6) ((- 2) (4 {((- 2) + 3)}^{3}) + {((- 2) + 3)}^{4})$

Step 5.2.1.5

Multiply $7$ by $- 2$ .

$f'' (- 2) = - 14 + (- 14 + 6) ((- 2) (4 {((- 2) + 3)}^{3}) + {((- 2) + 3)}^{4})$

Step 5.2.1.6

Add $- 14$ and $6$ .

$f'' (- 2) = - 14 - 8 ((- 2) (4 {((- 2) + 3)}^{3}) + {((- 2) + 3)}^{4})$

Step 5.2.1.7

Simplify each term.

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Step 5.2.1.7.1

Add $- 2$ and $3$ .

$f'' (- 2) = - 14 - 8 (- 2 (4 \cdot 1^{3}) + {((- 2) + 3)}^{4})$

Step 5.2.1.7.2

One to any power is one.

$f'' (- 2) = - 14 - 8 (- 2 (4 \cdot 1) + {((- 2) + 3)}^{4})$

Step 5.2.1.7.3

Multiply $- 2 (4 \cdot 1)$ .

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Step 5.2.1.7.3.1

Multiply $4$ by $1$ .

$f'' (- 2) = - 14 - 8 (- 2 \cdot 4 + {((- 2) + 3)}^{4})$

Step 5.2.1.7.3.2

Multiply $- 2$ by $4$ .

$f'' (- 2) = - 14 - 8 (- 8 + {((- 2) + 3)}^{4})$

Step 5.2.1.7.4

Add $- 2$ and $3$ .

$f'' (- 2) = - 14 - 8 (- 8 + 1^{4})$

Step 5.2.1.7.5

One to any power is one.

$f'' (- 2) = - 14 - 8 (- 8 + 1)$

Step 5.2.1.8

Add $- 8$ and $1$ .

$f'' (- 2) = - 14 - 8 \cdot - 7$

Step 5.2.1.9

Multiply $- 8$ by $- 7$ .

$f'' (- 2) = - 14 + 56$

Step 5.2.2

Add $- 14$ and $56$ .

$f'' (- 2) = 42$

Step 5.2.3

The final answer is $42$ .

$42$

Step 5.3

The graph is concave up on the interval $(- 3, - \frac{6 + \sqrt{15}}{7})$ because $f'' (- 2)$ is positive.

Concave up on $(- 3, - \frac{6 + \sqrt{15}}{7})$ since $f'' (x)$ is positive

Step 6

Substitute any number from the interval $(- \frac{6 + \sqrt{15}}{7}, - \frac{6 - \sqrt{15}}{7})$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

Replace the variable $x$ with $- 1$ in the expression.

$f'' (- 1) = 7 (- 1) {((- 1) + 3)}^{4} + (7 (- 1) + 6) ((- 1) (4 {((- 1) + 3)}^{3}) + {((- 1) + 3)}^{4})$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Multiply $7$ by $- 1$ .

$f'' (- 1) = - 7 {((- 1) + 3)}^{4} + (7 (- 1) + 6) ((- 1) (4 {((- 1) + 3)}^{3}) + {((- 1) + 3)}^{4})$

Step 6.2.1.2

Add $- 1$ and $3$ .

$f'' (- 1) = - 7 \cdot 2^{4} + (7 (- 1) + 6) ((- 1) (4 {((- 1) + 3)}^{3}) + {((- 1) + 3)}^{4})$

Step 6.2.1.3

Raise $2$ to the power of $4$ .

$f'' (- 1) = - 7 \cdot 16 + (7 (- 1) + 6) ((- 1) (4 {((- 1) + 3)}^{3}) + {((- 1) + 3)}^{4})$

Step 6.2.1.4

Multiply $- 7$ by $16$ .

$f'' (- 1) = - 112 + (7 (- 1) + 6) ((- 1) (4 {((- 1) + 3)}^{3}) + {((- 1) + 3)}^{4})$

Step 6.2.1.5

Multiply $7$ by $- 1$ .

$f'' (- 1) = - 112 + (- 7 + 6) ((- 1) (4 {((- 1) + 3)}^{3}) + {((- 1) + 3)}^{4})$

Step 6.2.1.6

Add $- 7$ and $6$ .

$f'' (- 1) = - 112 - 1 ((- 1) (4 {((- 1) + 3)}^{3}) + {((- 1) + 3)}^{4})$

Step 6.2.1.7

Simplify each term.

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Step 6.2.1.7.1

Add $- 1$ and $3$ .

$f'' (- 1) = - 112 - 1 (- 1 (4 \cdot 2^{3}) + {((- 1) + 3)}^{4})$

Step 6.2.1.7.2

Raise $2$ to the power of $3$ .

$f'' (- 1) = - 112 - 1 (- 1 (4 \cdot 8) + {((- 1) + 3)}^{4})$

Step 6.2.1.7.3

Multiply $- 1 (4 \cdot 8)$ .

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Step 6.2.1.7.3.1

Multiply $4$ by $8$ .

$f'' (- 1) = - 112 - 1 (- 1 \cdot 32 + {((- 1) + 3)}^{4})$

Step 6.2.1.7.3.2

Multiply $- 1$ by $32$ .

$f'' (- 1) = - 112 - 1 (- 32 + {((- 1) + 3)}^{4})$

Step 6.2.1.7.4

Add $- 1$ and $3$ .

$f'' (- 1) = - 112 - 1 (- 32 + 2^{4})$

Step 6.2.1.7.5

Raise $2$ to the power of $4$ .

$f'' (- 1) = - 112 - 1 (- 32 + 16)$

Step 6.2.1.8

Add $- 32$ and $16$ .

$f'' (- 1) = - 112 - 1 \cdot - 16$

Step 6.2.1.9

Multiply $- 1$ by $- 16$ .

$f'' (- 1) = - 112 + 16$

Step 6.2.2

Add $- 112$ and $16$ .

$f'' (- 1) = - 96$

Step 6.2.3

The final answer is $- 96$ .

$- 96$

Step 6.3

The graph is concave down on the interval $(- \frac{6 + \sqrt{15}}{7}, - \frac{6 - \sqrt{15}}{7})$ because $f'' (- 1)$ is negative.

Concave down on $(- \frac{6 + \sqrt{15}}{7}, - \frac{6 - \sqrt{15}}{7})$ since $f'' (x)$ is negative

Step 7

Substitute any number from the interval $(- \frac{6 - \sqrt{15}}{7}, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 7.1

Replace the variable $x$ with $0$ in the expression.

$f'' (0) = 7 (0) {((0) + 3)}^{4} + (7 (0) + 6) ((0) (4 {((0) + 3)}^{3}) + {((0) + 3)}^{4})$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Multiply $7$ by $0$ .

$f'' (0) = 0 {((0) + 3)}^{4} + (7 (0) + 6) ((0) (4 {((0) + 3)}^{3}) + {((0) + 3)}^{4})$

Step 7.2.1.2

Add $0$ and $3$ .

$f'' (0) = 0 \cdot 3^{4} + (7 (0) + 6) ((0) (4 {((0) + 3)}^{3}) + {((0) + 3)}^{4})$

Step 7.2.1.3

Raise $3$ to the power of $4$ .

$f'' (0) = 0 \cdot 81 + (7 (0) + 6) ((0) (4 {((0) + 3)}^{3}) + {((0) + 3)}^{4})$

Step 7.2.1.4

Multiply $0$ by $81$ .

$f'' (0) = 0 + (7 (0) + 6) ((0) (4 {((0) + 3)}^{3}) + {((0) + 3)}^{4})$

Step 7.2.1.5

Multiply $7$ by $0$ .

$f'' (0) = 0 + (0 + 6) ((0) (4 {((0) + 3)}^{3}) + {((0) + 3)}^{4})$

Step 7.2.1.6

Add $0$ and $6$ .

$f'' (0) = 0 + 6 ((0) (4 {((0) + 3)}^{3}) + {((0) + 3)}^{4})$

Step 7.2.1.7

Simplify each term.

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Step 7.2.1.7.1

Add $0$ and $3$ .

$f'' (0) = 0 + 6 (0 (4 \cdot 3^{3}) + {((0) + 3)}^{4})$

Step 7.2.1.7.2

Raise $3$ to the power of $3$ .

$f'' (0) = 0 + 6 (0 (4 \cdot 27) + {((0) + 3)}^{4})$

Step 7.2.1.7.3

Multiply $0 (4 \cdot 27)$ .

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Step 7.2.1.7.3.1

Multiply $4$ by $27$ .

$f'' (0) = 0 + 6 (0 \cdot 108 + {((0) + 3)}^{4})$

Step 7.2.1.7.3.2

Multiply $0$ by $108$ .

$f'' (0) = 0 + 6 (0 + {((0) + 3)}^{4})$

Step 7.2.1.7.4

Add $0$ and $3$ .

$f'' (0) = 0 + 6 (0 + 3^{4})$

Step 7.2.1.7.5

Raise $3$ to the power of $4$ .

$f'' (0) = 0 + 6 (0 + 81)$

Step 7.2.1.8

Add $0$ and $81$ .

$f'' (0) = 0 + 6 \cdot 81$

Step 7.2.1.9

Multiply $6$ by $81$ .

$f'' (0) = 0 + 486$

Step 7.2.2

Add $0$ and $486$ .

$f'' (0) = 486$

Step 7.2.3

The final answer is $486$ .

$486$

Step 7.3

The graph is concave up on the interval $(- \frac{6 - \sqrt{15}}{7}, \infty)$ because $f'' (0)$ is positive.

Concave up on $(- \frac{6 - \sqrt{15}}{7}, \infty)$ since $f'' (x)$ is positive

Step 8

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave down on $(- \infty, - 3)$ since $f'' (x)$ is negative

Concave up on $(- 3, - \frac{6 + \sqrt{15}}{7})$ since $f'' (x)$ is positive

Concave down on $(- \frac{6 + \sqrt{15}}{7}, - \frac{6 - \sqrt{15}}{7})$ since $f'' (x)$ is negative

Concave up on $(- \frac{6 - \sqrt{15}}{7}, \infty)$ since $f'' (x)$ is positive

Step 9