Calculus Examples

$f (x) = 10 x^{3} - 3 x^{5}$

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

By the Sum Rule, the derivative of $10 x^{3} - 3 x^{5}$ with respect to $x$ is $\frac{d}{d x} [10 x^{3}] + \frac{d}{d x} [- 3 x^{5}]$ .

$f' (x) = \frac{d}{d x} (10 x^{3}) + \frac{d}{d x} (- 3 x^{5})$

Step 1.1.1.2

Evaluate $\frac{d}{d x} [10 x^{3}]$ .

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Step 1.1.1.2.1

Since $10$ is constant with respect to $x$ , the derivative of $10 x^{3}$ with respect to $x$ is $10 \frac{d}{d x} [x^{3}]$ .

$f' (x) = 10 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 3 x^{5})$

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 $n = 3$ .

$f' (x) = 10 (3 x^{2}) + \frac{d}{d x} (- 3 x^{5})$

Step 1.1.1.2.3

Multiply $3$ by $10$ .

$f' (x) = 30 x^{2} + \frac{d}{d x} (- 3 x^{5})$

Step 1.1.1.3

Evaluate $\frac{d}{d x} [- 3 x^{5}]$ .

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Step 1.1.1.3.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x^{5}$ with respect to $x$ is $- 3 \frac{d}{d x} [x^{5}]$ .

$f' (x) = 30 x^{2} - 3 \frac{d}{d x} x^{5}$

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 $n = 5$ .

$f' (x) = 30 x^{2} - 3 (5 x^{4})$

Step 1.1.1.3.3

Multiply $5$ by $- 3$ .

$f' (x) = 30 x^{2} - 15 x^{4}$

Step 1.1.1.4

Reorder terms.

$f' (x) = - 15 x^{4} + 30 x^{2}$

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

By the Sum Rule, the derivative of $- 15 x^{4} + 30 x^{2}$ with respect to $x$ is $\frac{d}{d x} [- 15 x^{4}] + \frac{d}{d x} [30 x^{2}]$ .

$f'' (x) = \frac{d}{d x} (- 15 x^{4}) + \frac{d}{d x} (30 x^{2})$

Step 1.1.2.2

Evaluate $\frac{d}{d x} [- 15 x^{4}]$ .

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Step 1.1.2.2.1

Since $- 15$ is constant with respect to $x$ , the derivative of $- 15 x^{4}$ with respect to $x$ is $- 15 \frac{d}{d x} [x^{4}]$ .

$f'' (x) = - 15 \frac{d}{d x} x^{4} + \frac{d}{d x} (30 x^{2})$

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 $n = 4$ .

$f'' (x) = - 15 (4 x^{3}) + \frac{d}{d x} (30 x^{2})$

Step 1.1.2.2.3

Multiply $4$ by $- 15$ .

$f'' (x) = - 60 x^{3} + \frac{d}{d x} (30 x^{2})$

Step 1.1.2.3

Evaluate $\frac{d}{d x} [30 x^{2}]$ .

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Step 1.1.2.3.1

Since $30$ is constant with respect to $x$ , the derivative of $30 x^{2}$ with respect to $x$ is $30 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = - 60 x^{3} + 30 \frac{d}{d x} (x^{2})$

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 $n = 2$ .

$f'' (x) = - 60 x^{3} + 30 (2 x)$

Step 1.1.2.3.3

Multiply $2$ by $30$ .

$f'' (x) = - 60 x^{3} + 60 x$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $- 60 x^{3} + 60 x$ .

$- 60 x^{3} + 60 x$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $- 60 x^{3} + 60 x = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$- 60 x^{3} + 60 x = 0$

Step 1.2.2

Factor the left side of the equation.

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Step 1.2.2.1

Factor $- 60 x$ out of $- 60 x^{3} + 60 x$ .

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Step 1.2.2.1.1

Factor $- 60 x$ out of $- 60 x^{3}$ .

$- 60 x \cdot x^{2} + 60 x = 0$

Step 1.2.2.1.2

Factor $- 60 x$ out of $60 x$ .

$- 60 x \cdot x^{2} - 60 x \cdot - 1 = 0$

Step 1.2.2.1.3

Factor $- 60 x$ out of $- 60 x (x^{2}) - 60 x (- 1)$ .

$- 60 x (x^{2} - 1) = 0$

Step 1.2.2.2

Rewrite $1$ as $1^{2}$ .

$- 60 x (x^{2} - 1^{2}) = 0$

Step 1.2.2.3

Factor.

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Step 1.2.2.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$- 60 x ((x + 1) (x - 1)) = 0$

Step 1.2.2.3.2

Remove unnecessary parentheses.

$- 60 x (x + 1) (x - 1) = 0$

Step 1.2.3

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

$x = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 1.2.4

Set $x$ equal to $0$ .

$x = 0$

Step 1.2.5

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

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Step 1.2.5.1

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

$x + 1 = 0$

Step 1.2.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 1.2.6

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

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Step 1.2.6.1

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

$x - 1 = 0$

Step 1.2.6.2

Add $1$ to both sides of the equation.

$x = 1$

Step 1.2.7

The final solution is all the values that make $- 60 x (x + 1) (x - 1) = 0$ true.

$x = 0, - 1, 1$

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.

$(- \infty, - 1) \cup (- 1, 0) \cup (0, 1) \cup (1, \infty)$

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 $- 4$ in the expression.

$f'' (- 4) = - 60 {(- 4)}^{3} + 60 (- 4)$

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

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

$f'' (- 4) = - 60 \cdot - 64 + 60 (- 4)$

Step 4.2.1.2

Multiply $- 60$ by $- 64$ .

$f'' (- 4) = 3840 + 60 (- 4)$

Step 4.2.1.3

Multiply $60$ by $- 4$ .

$f'' (- 4) = 3840 - 240$

Step 4.2.2

Subtract $240$ from $3840$ .

$f'' (- 4) = 3600$

Step 4.2.3

The final answer is $3600$ .

$3600$

Step 4.3

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

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

Step 5

Substitute any number from the interval $(- 1, 0)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

Replace the variable $x$ with $- 0.5$ in the expression.

$f'' (- 0.5) = - 60 {(- 0.5)}^{3} + 60 (- 0.5)$

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 $- 0.5$ to the power of $3$ .

$f'' (- 0.5) = - 60 \cdot - 0.125 + 60 (- 0.5)$

Step 5.2.1.2

Multiply $- 60$ by $- 0.125$ .

$f'' (- 0.5) = 7.5 + 60 (- 0.5)$

Step 5.2.1.3

Multiply $60$ by $- 0.5$ .

$f'' (- 0.5) = 7.5 - 30$

Step 5.2.2

Subtract $30$ from $7.5$ .

$f'' (- 0.5) = - 22.5$

Step 5.2.3

The final answer is $- 22.5$ .

$- 22.5$

Step 5.3

The graph is concave down on the interval $(- 1, 0)$ because $f'' (- 0.5)$ is negative.

Concave down on $(- 1, 0)$ since $f'' (x)$ is negative

Step 6

Substitute any number from the interval $(0, 1)$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

Replace the variable $x$ with $0.5$ in the expression.

$f'' (0.5) = - 60 {(0.5)}^{3} + 60 (0.5)$

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 $0.5$ to the power of $3$ .

$f'' (0.5) = - 60 \cdot 0.125 + 60 (0.5)$

Step 6.2.1.2

Multiply $- 60$ by $0.125$ .

$f'' (0.5) = - 7.5 + 60 (0.5)$

Step 6.2.1.3

Multiply $60$ by $0.5$ .

$f'' (0.5) = - 7.5 + 30$

Step 6.2.2

Add $- 7.5$ and $30$ .

$f'' (0.5) = 22.5$

Step 6.2.3

The final answer is $22.5$ .

$22.5$

Step 6.3

The graph is concave up on the interval $(0, 1)$ because $f'' (0.5)$ is positive.

Concave up on $(0, 1)$ since $f'' (x)$ is positive

Step 7

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

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Step 7.1

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

$f'' (4) = - 60 {(4)}^{3} + 60 (4)$

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

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

$f'' (4) = - 60 \cdot 64 + 60 (4)$

Step 7.2.1.2

Multiply $- 60$ by $64$ .

$f'' (4) = - 3840 + 60 (4)$

Step 7.2.1.3

Multiply $60$ by $4$ .

$f'' (4) = - 3840 + 240$

Step 7.2.2

Add $- 3840$ and $240$ .

$f'' (4) = - 3600$

Step 7.2.3

The final answer is $- 3600$ .

$- 3600$

Step 7.3

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

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

Step 8

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

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

Concave down on $(- 1, 0)$ since $f'' (x)$ is negative

Concave up on $(0, 1)$ since $f'' (x)$ is positive

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

Step 9