Calculus Examples

$f (x) = x^{3} - 6 x^{2} - 63 x$

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 $x^{3} - 6 x^{2} - 63 x$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 6 x^{2}] + \frac{d}{d x} [- 63 x]$ .

$f' (x) = \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 6 x^{2}) + \frac{d}{d x} (- 63 x)$

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

$f' (x) = 3 x^{2} + \frac{d}{d x} (- 6 x^{2}) + \frac{d}{d x} (- 63 x)$

Evaluate $\frac{d}{d x} [- 6 x^{2}]$ .

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Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 x^{2}$ with respect to $x$ is $- 6 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 3 x^{2} - 6 \frac{d}{d x} x^{2} + \frac{d}{d x} (- 63 x)$

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

$f' (x) = 3 x^{2} - 6 (2 x) + \frac{d}{d x} (- 63 x)$

Multiply $2$ by $- 6$ .

$f' (x) = 3 x^{2} - 12 x + \frac{d}{d x} (- 63 x)$

Evaluate $\frac{d}{d x} [- 63 x]$ .

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Since $- 63$ is constant with respect to $x$ , the derivative of $- 63 x$ with respect to $x$ is $- 63 \frac{d}{d x} [x]$ .

$f' (x) = 3 x^{2} - 12 x - 63 \frac{d}{d x} x$

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

$f' (x) = 3 x^{2} - 12 x - 63 \cdot 1$

Multiply $- 63$ by $1$ .

$f' (x) = 3 x^{2} - 12 x - 63$

The first derivative of $f (x)$ with respect to $x$ is $3 x^{2} - 12 x - 63$ .

$3 x^{2} - 12 x - 63$

Step 2

Set the first derivative equal to $0$ then solve the equation $3 x^{2} - 12 x - 63 = 0$ .

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Set the first derivative equal to $0$ .

$3 x^{2} - 12 x - 63 = 0$

Factor the left side of the equation.

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Factor $3$ out of $3 x^{2} - 12 x - 63$ .

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Factor $3$ out of $3 x^{2}$ .

$3 (x^{2}) - 12 x - 63 = 0$

Factor $3$ out of $- 12 x$ .

$3 (x^{2}) + 3 (- 4 x) - 63 = 0$

Factor $3$ out of $- 63$ .

$3 x^{2} + 3 (- 4 x) + 3 \cdot - 21 = 0$

Factor $3$ out of $3 x^{2} + 3 (- 4 x)$ .

$3 (x^{2} - 4 x) + 3 \cdot - 21 = 0$

Factor $3$ out of $3 (x^{2} - 4 x) + 3 \cdot - 21$ .

$3 (x^{2} - 4 x - 21) = 0$

Factor.

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Factor $x^{2} - 4 x - 21$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 21$ and whose sum is $- 4$ .

$- 7, 3$

Write the factored form using these integers.

$3 ((x - 7) (x + 3)) = 0$

Remove unnecessary parentheses.

$3 (x - 7) (x + 3) = 0$

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

$x - 7 = 0$

$x + 3 = 0$

Set $x - 7$ equal to $0$ and solve for $x$ .

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Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Set $x + 3$ equal to $0$ and solve for $x$ .

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Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $3 (x - 7) (x + 3) = 0$ true.

$x = 7, - 3$

Step 3

The values which make the derivative equal to $0$ are $7, - 3$ .

$7, - 3$

Step 4

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

$(- \infty, - 3) \cup (- 3, 7) \cup (7, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 3)$ into the derivative to determine if the function is increasing or decreasing.

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

$f' (- 4) = 3 {(- 4)}^{2} - 12 \cdot - 4 - 63$

Simplify the result.

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Simplify each term.

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

$f' (- 4) = 3 \cdot 16 - 12 \cdot - 4 - 63$

Multiply $3$ by $16$ .

$f' (- 4) = 48 - 12 \cdot - 4 - 63$

Multiply $- 12$ by $- 4$ .

$f' (- 4) = 48 + 48 - 63$

Simplify by adding and subtracting.

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Add $48$ and $48$ .

$f' (- 4) = 96 - 63$

Subtract $63$ from $96$ .

$f' (- 4) = 33$

The final answer is $33$ .

$33$

At $x = - 4$ the derivative is $33$ . Since this is positive, the function is increasing on $(- \infty, - 3)$ .

Increasing on $(- \infty, - 3)$ since $f' (x) > 0$

Step 6

Substitute a value from the interval $(- 3, 7)$ into the derivative to determine if the function is increasing or decreasing.

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Replace the variable $x$ with $2$ in the expression.

$f' (2) = 3 {(2)}^{2} - 12 \cdot 2 - 63$

Simplify the result.

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Simplify each term.

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

$f' (2) = 3 \cdot 4 - 12 \cdot 2 - 63$

Multiply $3$ by $4$ .

$f' (2) = 12 - 12 \cdot 2 - 63$

Multiply $- 12$ by $2$ .

$f' (2) = 12 - 24 - 63$

Simplify by subtracting numbers.

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Subtract $24$ from $12$ .

$f' (2) = - 12 - 63$

Subtract $63$ from $- 12$ .

$f' (2) = - 75$

The final answer is $- 75$ .

$- 75$

At $x = 2$ the derivative is $- 75$ . Since this is negative, the function is decreasing on $(- 3, 7)$ .

Decreasing on $(- 3, 7)$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(7, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Replace the variable $x$ with $8$ in the expression.

$f' (8) = 3 {(8)}^{2} - 12 \cdot 8 - 63$

Simplify the result.

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Simplify each term.

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Raise $8$ to the power of $2$ .

$f' (8) = 3 \cdot 64 - 12 \cdot 8 - 63$

Multiply $3$ by $64$ .

$f' (8) = 192 - 12 \cdot 8 - 63$

Multiply $- 12$ by $8$ .

$f' (8) = 192 - 96 - 63$

Simplify by subtracting numbers.

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Subtract $96$ from $192$ .

$f' (8) = 96 - 63$

Subtract $63$ from $96$ .

$f' (8) = 33$

The final answer is $33$ .

$33$

At $x = 8$ the derivative is $33$ . Since this is positive, the function is increasing on $(7, \infty)$ .

Increasing on $(7, \infty)$ since $f' (x) > 0$

Step 8

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

Increasing on: $(- \infty, - 3), (7, \infty)$

Decreasing on: $(- 3, 7)$

Step 9