Calculus Examples

$f (x) = x^{4} - 72 x^{2}$

Step 1

Find the first derivative.

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Find the first derivative.

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Differentiate.

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By the Sum Rule, the derivative of $x^{4} - 72 x^{2}$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 72 x^{2}]$ .

$f' (x) = \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 72 x^{2})$

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

$f' (x) = 4 x^{3} + \frac{d}{d x} (- 72 x^{2})$

Evaluate $\frac{d}{d x} [- 72 x^{2}]$ .

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Since $- 72$ is constant with respect to $x$ , the derivative of $- 72 x^{2}$ with respect to $x$ is $- 72 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 4 x^{3} - 72 \frac{d}{d x} x^{2}$

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

$f' (x) = 4 x^{3} - 72 (2 x)$

Multiply $2$ by $- 72$ .

$f' (x) = 4 x^{3} - 144 x$

The first derivative of $f (x)$ with respect to $x$ is $4 x^{3} - 144 x$ .

$4 x^{3} - 144 x$

Step 2

Set the first derivative equal to $0$ then solve the equation $4 x^{3} - 144 x = 0$ .

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Set the first derivative equal to $0$ .

$4 x^{3} - 144 x = 0$

Factor the left side of the equation.

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Factor $4 x$ out of $4 x^{3} - 144 x$ .

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Factor $4 x$ out of $4 x^{3}$ .

$4 x (x^{2}) - 144 x = 0$

Factor $4 x$ out of $- 144 x$ .

$4 x (x^{2}) + 4 x (- 36) = 0$

Factor $4 x$ out of $4 x (x^{2}) + 4 x (- 36)$ .

$4 x (x^{2} - 36) = 0$

Rewrite $36$ as $6^{2}$ .

$4 x (x^{2} - 6^{2}) = 0$

Factor.

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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 $b = 6$ .

$4 x ((x + 6) (x - 6)) = 0$

Remove unnecessary parentheses.

$4 x (x + 6) (x - 6) = 0$

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

$x = 0$

$x + 6 = 0$

$x - 6 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 6$ equal to $0$ and solve for $x$ .

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Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Set $x - 6$ equal to $0$ and solve for $x$ .

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Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

The final solution is all the values that make $4 x (x + 6) (x - 6) = 0$ true.

$x = 0, - 6, 6$

Step 3

The values which make the derivative equal to $0$ are $0, - 6, 6$ .

$0, - 6, 6$

Step 4

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the derivative $0$ or undefined.

$(- \infty, - 6) \cup (- 6, 0) \cup (0, 6) \cup (6, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 6)$ into the derivative to determine if the function is increasing or decreasing.

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Replace the variable $x$ with $- 7$ in the expression.

$f' (- 7) = 4 {(- 7)}^{3} - 144 \cdot - 7$

Simplify the result.

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Simplify each term.

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Raise $- 7$ to the power of $3$ .

$f' (- 7) = 4 \cdot - 343 - 144 \cdot - 7$

Multiply $4$ by $- 343$ .

$f' (- 7) = - 1372 - 144 \cdot - 7$

Multiply $- 144$ by $- 7$ .

$f' (- 7) = - 1372 + 1008$

Add $- 1372$ and $1008$ .

$f' (- 7) = - 364$

The final answer is $- 364$ .

$- 364$

At $x = - 7$ the derivative is $- 364$ . Since this is negative, the function is decreasing on $(- \infty, - 6)$ .

Decreasing on $(- \infty, - 6)$ since $f' (x) < 0$

Step 6

Substitute a value from the interval $(- 6, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Replace the variable $x$ with $- 3$ in the expression.

$f' (- 3) = 4 {(- 3)}^{3} - 144 \cdot - 3$

Simplify the result.

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Simplify each term.

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

$f' (- 3) = 4 \cdot - 27 - 144 \cdot - 3$

Multiply $4$ by $- 27$ .

$f' (- 3) = - 108 - 144 \cdot - 3$

Multiply $- 144$ by $- 3$ .

$f' (- 3) = - 108 + 432$

Add $- 108$ and $432$ .

$f' (- 3) = 324$

The final answer is $324$ .

$324$

At $x = - 3$ the derivative is $324$ . Since this is positive, the function is increasing on $(- 6, 0)$ .

Increasing on $(- 6, 0)$ since $f' (x) > 0$

Step 7

Substitute a value from the interval $(0, 6)$ into the derivative to determine if the function is increasing or decreasing.

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Replace the variable $x$ with $3$ in the expression.

$f' (3) = 4 {(3)}^{3} - 144 \cdot 3$

Simplify the result.

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Simplify each term.

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

$f' (3) = 4 \cdot 27 - 144 \cdot 3$

Multiply $4$ by $27$ .

$f' (3) = 108 - 144 \cdot 3$

Multiply $- 144$ by $3$ .

$f' (3) = 108 - 432$

Subtract $432$ from $108$ .

$f' (3) = - 324$

The final answer is $- 324$ .

$- 324$

At $x = 3$ the derivative is $- 324$ . Since this is negative, the function is decreasing on $(0, 6)$ .

Decreasing on $(0, 6)$ since $f' (x) < 0$

Step 8

Substitute a value from the interval $(6, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Replace the variable $x$ with $7$ in the expression.

$f' (7) = 4 {(7)}^{3} - 144 \cdot 7$

Simplify the result.

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Simplify each term.

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Raise $7$ to the power of $3$ .

$f' (7) = 4 \cdot 343 - 144 \cdot 7$

Multiply $4$ by $343$ .

$f' (7) = 1372 - 144 \cdot 7$

Multiply $- 144$ by $7$ .

$f' (7) = 1372 - 1008$

Subtract $1008$ from $1372$ .

$f' (7) = 364$

The final answer is $364$ .

$364$

At $x = 7$ the derivative is $364$ . Since this is positive, the function is increasing on $(6, \infty)$ .

Increasing on $(6, \infty)$ since $f' (x) > 0$

Step 9

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- 6, 0), (6, \infty)$

Decreasing on: $(- \infty, - 6), (0, 6)$

Step 10