Calculus Examples

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

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.

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Step 1.1.1.1.1

By the Sum Rule, the derivative of $x^{3} - 3 x^{2} - 4 x + 5$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [5]$ .

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

Step 1.1.1.1.2

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

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

Step 1.1.1.2

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

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Step 1.1.1.2.1

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

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

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 $n = 2$ .

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

Step 1.1.1.2.3

Multiply $2$ by $- 3$ .

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

Step 1.1.1.3

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

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Step 1.1.1.3.1

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

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

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 $n = 1$ .

$3 x^{2} - 6 x - 4 \cdot 1 + \frac{d}{d x} [5]$

Step 1.1.1.3.3

Multiply $- 4$ by $1$ .

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

Step 1.1.1.4

Differentiate using the Constant Rule.

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Step 1.1.1.4.1

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

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

Step 1.1.1.4.2

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

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

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

$\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [- 4]$

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

$3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [- 4]$

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 $n = 2$ .

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

Step 1.1.2.2.3

Multiply $2$ by $3$ .

$6 x + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [- 4]$

Step 1.1.2.3

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

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Step 1.1.2.3.1

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

$6 x - 6 \frac{d}{d x} [x] + \frac{d}{d x} [- 4]$

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 $n = 1$ .

$6 x - 6 \cdot 1 + \frac{d}{d x} [- 4]$

Step 1.1.2.3.3

Multiply $- 6$ by $1$ .

$6 x - 6 + \frac{d}{d x} [- 4]$

Step 1.1.2.4

Differentiate using the Constant Rule.

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Step 1.1.2.4.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$6 x - 6 + 0$

Step 1.1.2.4.2

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

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

Step 1.1.3

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

$6 x - 6$

Step 1.2

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

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Step 1.2.1

Set the second derivative equal to $0$ .

$6 x - 6 = 0$

Step 1.2.2

Add $6$ to both sides of the equation.

$6 x = 6$

Step 1.2.3

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

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Step 1.2.3.1

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

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

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.

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

Step 1.2.3.2.1.2

Divide $x$ by $1$ .

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

Step 1.2.3.3

Simplify the right side.

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Step 1.2.3.3.1

Divide $6$ by $6$ .

$x = 1$

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.

$(- \infty, 1) \cup (1, \infty)$

Step 4

Substitute any number from the interval $(- \infty, 1)$ 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.

$f'' (0) = 6 (0) - 6$

Step 4.2

Simplify the result.

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Step 4.2.1

Multiply $6$ by $0$ .

$f'' (0) = 0 - 6$

Step 4.2.2

Subtract $6$ from $0$ .

$f'' (0) = - 6$

Step 4.2.3

The final answer is $- 6$ .

$- 6$

Step 4.3

The graph is concave down on the interval $(- \infty, 1)$ because $f'' (0)$ is negative.

Concave down on $(- \infty, 1)$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(1, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

Replace the variable $x$ with $4$ in the expression.

$f'' (4) = 6 (4) - 6$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $6$ by $4$ .

$f'' (4) = 24 - 6$

Step 5.2.2

Subtract $6$ from $24$ .

$f'' (4) = 18$

Step 5.2.3

The final answer is $18$ .

$18$

Step 5.3

The graph is concave up on the interval $(1, \infty)$ because $f'' (4)$ is positive.

Concave up on $(1, \infty)$ since $f'' (x)$ is positive

Step 6

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

Concave down on $(- \infty, 1)$ since $f'' (x)$ is negative

Concave up on $(1, \infty)$ since $f'' (x)$ is positive

Step 7