Calculus Examples

$f (x) = x^{4} + x^{3} - 3 x^{2} + 9$

Step 1

Find the second derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate.

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Step 1.1.1.1

By the Sum Rule, the derivative of $x^{4} + x^{3} - 3 x^{2} + 9$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [9]$ .

$f' (x) = \frac{d}{d x} (x^{4}) + \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 3 x^{2}) + \frac{d}{d x} (9)$

Step 1.1.1.2

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

$f' (x) = 4 x^{3} + \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 3 x^{2}) + \frac{d}{d x} (9)$

Step 1.1.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f' (x) = 4 x^{3} + 3 x^{2} + \frac{d}{d x} (- 3 x^{2}) + \frac{d}{d x} (9)$

Step 1.1.2

Evaluate $\frac{d}{d x} [- 3 x^{2}]$ .

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Step 1.1.2.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x^{2}$ with respect to $x$ is $- 3 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 4 x^{3} + 3 x^{2} - 3 \frac{d}{d x} x^{2} + \frac{d}{d x} (9)$

Step 1.1.2.2

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

$f' (x) = 4 x^{3} + 3 x^{2} - 3 (2 x) + \frac{d}{d x} (9)$

Step 1.1.2.3

Multiply $2$ by $- 3$ .

$f' (x) = 4 x^{3} + 3 x^{2} - 6 x + \frac{d}{d x} (9)$

Step 1.1.3

Differentiate using the Constant Rule.

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Step 1.1.3.1

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$f' (x) = 4 x^{3} + 3 x^{2} - 6 x + 0$

Step 1.1.3.2

Add $4 x^{3} + 3 x^{2} - 6 x$ and $0$ .

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

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $4 x^{3} + 3 x^{2} - 6 x$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x]$ .

$f'' (x) = \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 6 x)$

Step 1.2.2

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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Step 1.2.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3}$ with respect to $x$ is $4 \frac{d}{d x} [x^{3}]$ .

$f'' (x) = 4 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 6 x)$

Step 1.2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f'' (x) = 4 (3 x^{2}) + \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 6 x)$

Step 1.2.2.3

Multiply $3$ by $4$ .

$f'' (x) = 12 x^{2} + \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 6 x)$

Step 1.2.3

Evaluate $\frac{d}{d x} [3 x^{2}]$ .

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Step 1.2.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 12 x^{2} + 3 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 6 x)$

Step 1.2.3.2

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

$f'' (x) = 12 x^{2} + 3 (2 x) + \frac{d}{d x} (- 6 x)$

Step 1.2.3.3

Multiply $2$ by $3$ .

$f'' (x) = 12 x^{2} + 6 x + \frac{d}{d x} (- 6 x)$

Step 1.2.4

Evaluate $\frac{d}{d x} [- 6 x]$ .

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Step 1.2.4.1

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 x$ with respect to $x$ is $- 6 \frac{d}{d x} [x]$ .

$f'' (x) = 12 x^{2} + 6 x - 6 \frac{d}{d x} x$

Step 1.2.4.2

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

$f'' (x) = 12 x^{2} + 6 x - 6 \cdot 1$

Step 1.2.4.3

Multiply $- 6$ by $1$ .

$f'' (x) = 12 x^{2} + 6 x - 6$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $12 x^{2} + 6 x - 6$ .

$12 x^{2} + 6 x - 6$

Step 2

Set the second derivative equal to $0$ then solve the equation $12 x^{2} + 6 x - 6 = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$12 x^{2} + 6 x - 6 = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $6$ out of $12 x^{2} + 6 x - 6$ .

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Step 2.2.1.1

Factor $6$ out of $12 x^{2}$ .

$6 (2 x^{2}) + 6 x - 6 = 0$

Step 2.2.1.2

Factor $6$ out of $6 x$ .

$6 (2 x^{2}) + 6 (x) - 6 = 0$

Step 2.2.1.3

Factor $6$ out of $- 6$ .

$6 (2 x^{2}) + 6 (x) + 6 (- 1) = 0$

Step 2.2.1.4

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

$6 (2 x^{2} + x) + 6 (- 1) = 0$

Step 2.2.1.5

Factor $6$ out of $6 (2 x^{2} + x) + 6 (- 1)$ .

$6 (2 x^{2} + x - 1) = 0$

Step 2.2.2

Factor.

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Step 2.2.2.1

Factor by grouping.

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Step 2.2.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = 1$ .

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Step 2.2.2.1.1.1

Multiply by $1$ .

$6 (2 x^{2} + 1 x - 1) = 0$

Step 2.2.2.1.1.2

Rewrite $1$ as $- 1$ plus $2$

$6 (2 x^{2} + (- 1 + 2) x - 1) = 0$

Step 2.2.2.1.1.3

Apply the distributive property.

$6 (2 x^{2} - 1 x + 2 x - 1) = 0$

Step 2.2.2.1.2

Factor out the greatest common factor from each group.

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Step 2.2.2.1.2.1

Group the first two terms and the last two terms.

$6 ((2 x^{2} - 1 x) + 2 x - 1) = 0$

Step 2.2.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$6 (x (2 x - 1) + 1 (2 x - 1)) = 0$

Step 2.2.2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$6 ((2 x - 1) (x + 1)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$6 (2 x - 1) (x + 1) = 0$

Step 2.3

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

$2 x - 1 = 0$

$x + 1 = 0$

Step 2.4

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 2.4.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 2.4.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 2.4.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 2.4.2.2.1

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 2.5

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 2.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.6

The final solution is all the values that make $6 (2 x - 1) (x + 1) = 0$ true.

$x = \frac{1}{2}, - 1$

Step 3

Find the points where the second derivative is $0$ .

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Step 3.1

Substitute $\frac{1}{2}$ in $f (x) = x^{4} + x^{3} - 3 x^{2} + 9$ to find the value of $y$ .

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Step 3.1.1

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

$f (\frac{1}{2}) = {(\frac{1}{2})}^{4} + {(\frac{1}{2})}^{3} - 3 {(\frac{1}{2})}^{2} + 9$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

Apply the product rule to $\frac{1}{2}$ .

$f (\frac{1}{2}) = \frac{1^{4}}{2^{4}} + {(\frac{1}{2})}^{3} - 3 {(\frac{1}{2})}^{2} + 9$

Step 3.1.2.1.2

One to any power is one.

$f (\frac{1}{2}) = \frac{1}{2^{4}} + {(\frac{1}{2})}^{3} - 3 {(\frac{1}{2})}^{2} + 9$

Step 3.1.2.1.3

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

$f (\frac{1}{2}) = \frac{1}{16} + {(\frac{1}{2})}^{3} - 3 {(\frac{1}{2})}^{2} + 9$

Step 3.1.2.1.4

Apply the product rule to $\frac{1}{2}$ .

$f (\frac{1}{2}) = \frac{1}{16} + \frac{1^{3}}{2^{3}} - 3 {(\frac{1}{2})}^{2} + 9$

Step 3.1.2.1.5

One to any power is one.

$f (\frac{1}{2}) = \frac{1}{16} + \frac{1}{2^{3}} - 3 {(\frac{1}{2})}^{2} + 9$

Step 3.1.2.1.6

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

$f (\frac{1}{2}) = \frac{1}{16} + \frac{1}{8} - 3 {(\frac{1}{2})}^{2} + 9$

Step 3.1.2.1.7

Apply the product rule to $\frac{1}{2}$ .

$f (\frac{1}{2}) = \frac{1}{16} + \frac{1}{8} - 3 \frac{1^{2}}{2^{2}} + 9$

Step 3.1.2.1.8

One to any power is one.

$f (\frac{1}{2}) = \frac{1}{16} + \frac{1}{8} - 3 \frac{1}{2^{2}} + 9$

Step 3.1.2.1.9

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

$f (\frac{1}{2}) = \frac{1}{16} + \frac{1}{8} - 3 (\frac{1}{4}) + 9$

Step 3.1.2.1.10

Combine $- 3$ and $\frac{1}{4}$ .

$f (\frac{1}{2}) = \frac{1}{16} + \frac{1}{8} + \frac{- 3}{4} + 9$

Step 3.1.2.1.11

Move the negative in front of the fraction.

$f (\frac{1}{2}) = \frac{1}{16} + \frac{1}{8} - \frac{3}{4} + 9$

Step 3.1.2.2

Find the common denominator.

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Step 3.1.2.2.1

Multiply $\frac{1}{8}$ by $\frac{2}{2}$ .

$f (\frac{1}{2}) = \frac{1}{16} + \frac{1}{8} \cdot \frac{2}{2} - \frac{3}{4} + 9$

Step 3.1.2.2.2

Multiply $\frac{1}{8}$ by $\frac{2}{2}$ .

$f (\frac{1}{2}) = \frac{1}{16} + \frac{2}{8 \cdot 2} - \frac{3}{4} + 9$

Step 3.1.2.2.3

Multiply $\frac{3}{4}$ by $\frac{4}{4}$ .

$f (\frac{1}{2}) = \frac{1}{16} + \frac{2}{8 \cdot 2} - (\frac{3}{4} \cdot \frac{4}{4}) + 9$

Step 3.1.2.2.4

Multiply $\frac{3}{4}$ by $\frac{4}{4}$ .

$f (\frac{1}{2}) = \frac{1}{16} + \frac{2}{8 \cdot 2} - \frac{3 \cdot 4}{4 \cdot 4} + 9$

Step 3.1.2.2.5

Write $9$ as a fraction with denominator $1$ .

$f (\frac{1}{2}) = \frac{1}{16} + \frac{2}{8 \cdot 2} - \frac{3 \cdot 4}{4 \cdot 4} + \frac{9}{1}$

Step 3.1.2.2.6

Multiply $\frac{9}{1}$ by $\frac{16}{16}$ .

$f (\frac{1}{2}) = \frac{1}{16} + \frac{2}{8 \cdot 2} - \frac{3 \cdot 4}{4 \cdot 4} + \frac{9}{1} \cdot \frac{16}{16}$

Step 3.1.2.2.7

Multiply $\frac{9}{1}$ by $\frac{16}{16}$ .

$f (\frac{1}{2}) = \frac{1}{16} + \frac{2}{8 \cdot 2} - \frac{3 \cdot 4}{4 \cdot 4} + \frac{9 \cdot 16}{16}$

Step 3.1.2.2.8

Reorder the factors of $8 \cdot 2$ .

$f (\frac{1}{2}) = \frac{1}{16} + \frac{2}{2 \cdot 8} - \frac{3 \cdot 4}{4 \cdot 4} + \frac{9 \cdot 16}{16}$

Step 3.1.2.2.9

Multiply $2$ by $8$ .

$f (\frac{1}{2}) = \frac{1}{16} + \frac{2}{16} - \frac{3 \cdot 4}{4 \cdot 4} + \frac{9 \cdot 16}{16}$

Step 3.1.2.2.10

Multiply $4$ by $4$ .

$f (\frac{1}{2}) = \frac{1}{16} + \frac{2}{16} - \frac{3 \cdot 4}{16} + \frac{9 \cdot 16}{16}$

Step 3.1.2.3

Combine the numerators over the common denominator.

$f (\frac{1}{2}) = \frac{1 + 2 - 3 \cdot 4 + 9 \cdot 16}{16}$

Step 3.1.2.4

Simplify each term.

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Step 3.1.2.4.1

Multiply $- 3$ by $4$ .

$f (\frac{1}{2}) = \frac{1 + 2 - 12 + 9 \cdot 16}{16}$

Step 3.1.2.4.2

Multiply $9$ by $16$ .

$f (\frac{1}{2}) = \frac{1 + 2 - 12 + 144}{16}$

Step 3.1.2.5

Simplify by adding and subtracting.

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Step 3.1.2.5.1

Add $1$ and $2$ .

$f (\frac{1}{2}) = \frac{3 - 12 + 144}{16}$

Step 3.1.2.5.2

Subtract $12$ from $3$ .

$f (\frac{1}{2}) = \frac{- 9 + 144}{16}$

Step 3.1.2.5.3

Add $- 9$ and $144$ .

$f (\frac{1}{2}) = \frac{135}{16}$

Step 3.1.2.6

The final answer is $\frac{135}{16}$ .

$\frac{135}{16}$

Step 3.2

The point found by substituting $\frac{1}{2}$ in $f (x) = x^{4} + x^{3} - 3 x^{2} + 9$ is $(\frac{1}{2}, \frac{135}{16})$ . This point can be an inflection point.

$(\frac{1}{2}, \frac{135}{16})$

Step 3.3

Substitute $- 1$ in $f (x) = x^{4} + x^{3} - 3 x^{2} + 9$ to find the value of $y$ .

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Step 3.3.1

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

$f (- 1) = {(- 1)}^{4} + {(- 1)}^{3} - 3 {(- 1)}^{2} + 9$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Raise $- 1$ to the power of $4$ .

$f (- 1) = 1 + {(- 1)}^{3} - 3 {(- 1)}^{2} + 9$

Step 3.3.2.1.2

Raise $- 1$ to the power of $3$ .

$f (- 1) = 1 - 1 - 3 {(- 1)}^{2} + 9$

Step 3.3.2.1.3

Raise $- 1$ to the power of $2$ .

$f (- 1) = 1 - 1 - 3 \cdot 1 + 9$

Step 3.3.2.1.4

Multiply $- 3$ by $1$ .

$f (- 1) = 1 - 1 - 3 + 9$

Step 3.3.2.2

Simplify by adding and subtracting.

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Step 3.3.2.2.1

Subtract $1$ from $1$ .

$f (- 1) = 0 - 3 + 9$

Step 3.3.2.2.2

Subtract $3$ from $0$ .

$f (- 1) = - 3 + 9$

Step 3.3.2.2.3

Add $- 3$ and $9$ .

$f (- 1) = 6$

Step 3.3.2.3

The final answer is $6$ .

$6$

Step 3.4

The point found by substituting $- 1$ in $f (x) = x^{4} + x^{3} - 3 x^{2} + 9$ is $(- 1, 6)$ . This point can be an inflection point.

$(- 1, 6)$

Step 3.5

Determine the points that could be inflection points.

$(\frac{1}{2}, \frac{135}{16}), (- 1, 6)$

Step 4

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, - 1) \cup (- 1, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 1)$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 1.1$ in the expression.

$f'' (- 1.1) = 12 {(- 1.1)}^{2} + 6 (- 1.1) - 6$

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

Raise $- 1.1$ to the power of $2$ .

$f'' (- 1.1) = 12 \cdot 1.21 + 6 (- 1.1) - 6$

Step 5.2.1.2

Multiply $12$ by $1.21$ .

$f'' (- 1.1) = 14.52 + 6 (- 1.1) - 6$

Step 5.2.1.3

Multiply $6$ by $- 1.1$ .

$f'' (- 1.1) = 14.52 - 6.6 - 6$

Step 5.2.2

Simplify by subtracting numbers.

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Step 5.2.2.1

Subtract $6.6$ from $14.52$ .

$f'' (- 1.1) = 7.92 - 6$

Step 5.2.2.2

Subtract $6$ from $7.92$ .

$f'' (- 1.1) = 1.92$

Step 5.2.3

The final answer is $1.92$ .

$1.92$

Step 5.3

At $- 1.1$ , the second derivative is $1.92$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, - 1)$ .

Increasing on $(- \infty, - 1)$ since $f'' (x) > 0$

Step 6

Substitute a value from the interval $(- 1, \frac{1}{2})$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 0.25$ in the expression.

$f'' (- 0.25) = 12 {(- 0.25)}^{2} + 6 (- 0.25) - 6$

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

Raise $- 0.25$ to the power of $2$ .

$f'' (- 0.25) = 12 \cdot 0.0625 + 6 (- 0.25) - 6$

Step 6.2.1.2

Multiply $12$ by $0.0625$ .

$f'' (- 0.25) = 0.75 + 6 (- 0.25) - 6$

Step 6.2.1.3

Multiply $6$ by $- 0.25$ .

$f'' (- 0.25) = 0.75 - 1.5 - 6$

Step 6.2.2

Simplify by subtracting numbers.

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Step 6.2.2.1

Subtract $1.5$ from $0.75$ .

$f'' (- 0.25) = - 0.75 - 6$

Step 6.2.2.2

Subtract $6$ from $- 0.75$ .

$f'' (- 0.25) = - 6.75$

Step 6.2.3

The final answer is $- 6.75$ .

$- 6.75$

Step 6.3

At $- 0.25$ , the second derivative is $- 6.75$ . Since this is negative, the second derivative is decreasing on the interval $(- 1, \frac{1}{2})$

Decreasing on $(- 1, \frac{1}{2})$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(\frac{1}{2}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $0.6$ in the expression.

$f'' (0.6) = 12 {(0.6)}^{2} + 6 (0.6) - 6$

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

Raise $0.6$ to the power of $2$ .

$f'' (0.6) = 12 \cdot 0.36 + 6 (0.6) - 6$

Step 7.2.1.2

Multiply $12$ by $0.36$ .

$f'' (0.6) = 4.32 + 6 (0.6) - 6$

Step 7.2.1.3

Multiply $6$ by $0.6$ .

$f'' (0.6) = 4.32 + 3.6 - 6$

Step 7.2.2

Simplify by adding and subtracting.

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Step 7.2.2.1

Add $4.32$ and $3.6$ .

$f'' (0.6) = 7.92 - 6$

Step 7.2.2.2

Subtract $6$ from $7.92$ .

$f'' (0.6) = 1.92$

Step 7.2.3

The final answer is $1.92$ .

$1.92$

Step 7.3

At $0.6$ , the second derivative is $1.92$ . Since this is positive, the second derivative is increasing on the interval $(\frac{1}{2}, \infty)$ .

Increasing on $(\frac{1}{2}, \infty)$ since $f'' (x) > 0$

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection points in this case are $(- 1, 6), (\frac{1}{2}, \frac{135}{16})$ .

$(- 1, 6), (\frac{1}{2}, \frac{135}{16})$

Step 9