Calculus Examples

$f (x) = 2 - 15 x + 9 x^{2} - x^{3}$

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.

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Step 1.1.1.1.1

By the Sum Rule, the derivative of $2 - 15 x + 9 x^{2} - x^{3}$ with respect to $x$ is $\frac{d}{d x} [2] + \frac{d}{d x} [- 15 x] + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [- x^{3}]$ .

$\frac{d}{d x} [2] + \frac{d}{d x} [- 15 x] + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [- x^{3}]$

Step 1.1.1.1.2

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- 15 x] + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [- x^{3}]$

Step 1.1.1.2

Evaluate $\frac{d}{d x} [- 15 x]$ .

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Step 1.1.1.2.1

Since $- 15$ is constant with respect to $x$ , the derivative of $- 15 x$ with respect to $x$ is $- 15 \frac{d}{d x} [x]$ .

$0 - 15 \frac{d}{d x} [x] + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [- x^{3}]$

Step 1.1.1.2.2

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

$0 - 15 \cdot 1 + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [- x^{3}]$

Step 1.1.1.2.3

Multiply $- 15$ by $1$ .

$0 - 15 + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [- x^{3}]$

Step 1.1.1.3

Evaluate $\frac{d}{d x} [9 x^{2}]$ .

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Step 1.1.1.3.1

Since $9$ is constant with respect to $x$ , the derivative of $9 x^{2}$ with respect to $x$ is $9 \frac{d}{d x} [x^{2}]$ .

$0 - 15 + 9 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- x^{3}]$

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 $n = 2$ .

$0 - 15 + 9 (2 x) + \frac{d}{d x} [- x^{3}]$

Step 1.1.1.3.3

Multiply $2$ by $9$ .

$0 - 15 + 18 x + \frac{d}{d x} [- x^{3}]$

Step 1.1.1.4

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

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Step 1.1.1.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{3}$ with respect to $x$ is $- \frac{d}{d x} [x^{3}]$ .

$0 - 15 + 18 x - \frac{d}{d x} [x^{3}]$

Step 1.1.1.4.2

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

$0 - 15 + 18 x - (3 x^{2})$

Step 1.1.1.4.3

Multiply $3$ by $- 1$ .

$0 - 15 + 18 x - 3 x^{2}$

Step 1.1.1.5

Simplify.

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Step 1.1.1.5.1

Subtract $15$ from $0$ .

$- 15 + 18 x - 3 x^{2}$

Step 1.1.1.5.2

Reorder terms.

$f' (x) = - 3 x^{2} + 18 x - 15$

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

By the Sum Rule, the derivative of $- 3 x^{2} + 18 x - 15$ with respect to $x$ is $\frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [18 x] + \frac{d}{d x} [- 15]$ .

$\frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [18 x] + \frac{d}{d x} [- 15]$

Step 1.1.2.2

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

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Step 1.1.2.2.1

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

$- 3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [18 x] + \frac{d}{d x} [- 15]$

Step 1.1.2.2.2

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

$- 3 (2 x) + \frac{d}{d x} [18 x] + \frac{d}{d x} [- 15]$

Step 1.1.2.2.3

Multiply $2$ by $- 3$ .

$- 6 x + \frac{d}{d x} [18 x] + \frac{d}{d x} [- 15]$

Step 1.1.2.3

Evaluate $\frac{d}{d x} [18 x]$ .

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Step 1.1.2.3.1

Since $18$ is constant with respect to $x$ , the derivative of $18 x$ with respect to $x$ is $18 \frac{d}{d x} [x]$ .

$- 6 x + 18 \frac{d}{d x} [x] + \frac{d}{d x} [- 15]$

Step 1.1.2.3.2

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

$- 6 x + 18 \cdot 1 + \frac{d}{d x} [- 15]$

Step 1.1.2.3.3

Multiply $18$ by $1$ .

$- 6 x + 18 + \frac{d}{d x} [- 15]$

Step 1.1.2.4

Differentiate using the Constant Rule.

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Step 1.1.2.4.1

Since $- 15$ is constant with respect to $x$ , the derivative of $- 15$ with respect to $x$ is $0$ .

$- 6 x + 18 + 0$

Step 1.1.2.4.2

Add $- 6 x + 18$ and $0$ .

$f'' (x) = - 6 x + 18$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $- 6 x + 18$ .

$- 6 x + 18$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $- 6 x + 18 = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$- 6 x + 18 = 0$

Step 1.2.2

Subtract $18$ from both sides of the equation.

$- 6 x = - 18$

Step 1.2.3

Divide each term in $- 6 x = - 18$ by $- 6$ and simplify.

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Step 1.2.3.1

Divide each term in $- 6 x = - 18$ by $- 6$ .

$\frac{- 6 x}{- 6} = \frac{- 18}{- 6}$

Step 1.2.3.2

Simplify the left side.

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Step 1.2.3.2.1

Cancel the common factor of $- 6$ .

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Step 1.2.3.2.1.1

Cancel the common factor.

$\frac{- 6 x}{- 6} = \frac{- 18}{- 6}$

Step 1.2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 18}{- 6}$

Step 1.2.3.3

Simplify the right side.

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Step 1.2.3.3.1

Divide $- 18$ by $- 6$ .

$x = 3$

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.

$(- \infty, 3) \cup (3, \infty)$

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 $0$ in the expression.

$f'' (0) = - 6 \cdot 0 + 18$

Step 4.2

Simplify the result.

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Step 4.2.1

Multiply $- 6$ by $0$ .

$f'' (0) = 0 + 18$

Step 4.2.2

Add $0$ and $18$ .

$f'' (0) = 18$

Step 4.2.3

The final answer is $18$ .

$18$

Step 4.3

The graph is concave up on the interval $(- \infty, 3)$ because $f'' (0)$ is positive.

Concave up on $(- \infty, 3)$ since $f'' (x)$ is positive

Step 5

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

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Step 5.1

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

$f'' (6) = - 6 \cdot 6 + 18$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $- 6$ by $6$ .

$f'' (6) = - 36 + 18$

Step 5.2.2

Add $- 36$ and $18$ .

$f'' (6) = - 18$

Step 5.2.3

The final answer is $- 18$ .

$- 18$

Step 5.3

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

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

Step 6

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

Concave up on $(- \infty, 3)$ since $f'' (x)$ is positive

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

Step 7