Calculus Examples

$f (x) = \frac{3}{x - 7}$

Step 1

Find the first derivative.

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Find the first derivative.

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Differentiate using the Constant Multiple Rule.

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Since $3$ is constant with respect to $x$ , the derivative of $\frac{3}{x - 7}$ with respect to $x$ is $3 \frac{d}{d x} [\frac{1}{x - 7}]$ .

$f' (x) = 3 \frac{d}{d x} (\frac{1}{x - 7})$

Rewrite $\frac{1}{x - 7}$ as ${(x - 7)}^{- 1}$ .

$f' (x) = 3 \frac{d}{d x} ({(x - 7)}^{- 1})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x - 7$ .

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To apply the Chain Rule, set $u$ as $x - 7$ .

$f' (x) = 3 (\frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x - 7))$

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

$f' (x) = 3 (- u^{- 2} \frac{d}{d x} (x - 7))$

Replace all occurrences of $u$ with $x - 7$ .

$f' (x) = 3 (- {(x - 7)}^{- 2} \frac{d}{d x} (x - 7))$

Differentiate.

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Multiply $- 1$ by $3$ .

$f' (x) = - 3 ({(x - 7)}^{- 2} \frac{d}{d x} (x - 7))$

By the Sum Rule, the derivative of $x - 7$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 7]$ .

$f' (x) = - 3 {(x - 7)}^{- 2} (\frac{d}{d x} (x) + \frac{d}{d x} (- 7))$

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

$f' (x) = - 3 {(x - 7)}^{- 2} (1 + \frac{d}{d x} (- 7))$

Since $- 7$ is constant with respect to $x$ , the derivative of $- 7$ with respect to $x$ is $0$ .

$f' (x) = - 3 {(x - 7)}^{- 2} (1 + 0)$

Simplify the expression.

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Add $1$ and $0$ .

$f' (x) = - 3 {(x - 7)}^{- 2} \cdot 1$

Multiply $- 3$ by $1$ .

$f' (x) = - 3 {(x - 7)}^{- 2}$

Simplify.

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Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (x) = - 3 \frac{1}{{(x - 7)}^{2}}$

Combine terms.

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Combine $- 3$ and $\frac{1}{{(x - 7)}^{2}}$ .

$f' (x) = \frac{- 3}{{(x - 7)}^{2}}$

Move the negative in front of the fraction.

$f' (x) = - \frac{3}{{(x - 7)}^{2}}$

The first derivative of $f (x)$ with respect to $x$ is $- \frac{3}{{(x - 7)}^{2}}$ .

$- \frac{3}{{(x - 7)}^{2}}$

Step 2

Set the first derivative equal to $0$ then solve the equation $- \frac{3}{{(x - 7)}^{2}} = 0$ .

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Set the first derivative equal to $0$ .

$- \frac{3}{{(x - 7)}^{2}} = 0$

Set the numerator equal to zero.

$3 = 0$

Since $3 \neq 0$ , there are no solutions.

No solution

Step 3

There are no values of $x$ in the domain of the original problem where the derivative is $0$ or undefined.

No critical points found

Step 4

Find where the derivative is undefined.

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Set the denominator in $\frac{3}{{(x - 7)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(x - 7)}^{2} = 0$

Solve for $x$ .

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Step 5

After finding the point that makes the derivative $f' (x) = - \frac{3}{{(x - 7)}^{2}}$ equal to $0$ or undefined, the interval to check where $f (x) = \frac{3}{x - 7}$ is increasing and where it is decreasing is $(- \infty, 7) \cup (7, \infty)$ .

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

Step 6

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

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

$f' (6) = - \frac{3}{{((6) - 7)}^{2}}$

Simplify the result.

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Simplify the denominator.

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Subtract $7$ from $6$ .

$f' (6) = - \frac{3}{{(- 1)}^{2}}$

Raise $- 1$ to the power of $2$ .

$f' (6) = - \frac{3}{1}$

Simplify the expression.

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Divide $3$ by $1$ .

$f' (6) = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$f' (6) = - 3$

The final answer is $- 3$ .

$- 3$

At $x = 6$ the derivative is $- 3$ . Since this is negative, the function is decreasing on $(- \infty, 7)$ .

Decreasing on $(- \infty, 7)$ 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.

$f' (8) = - \frac{3}{{((8) - 7)}^{2}}$

Simplify the result.

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Simplify the denominator.

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Subtract $7$ from $8$ .

$f' (8) = - \frac{3}{1^{2}}$

One to any power is one.

$f' (8) = - \frac{3}{1}$

Simplify the expression.

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Divide $3$ by $1$ .

$f' (8) = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$f' (8) = - 3$

The final answer is $- 3$ .

$- 3$

At $x = 8$ the derivative is $- 3$ . Since this is negative, the function is decreasing on $(7, \infty)$ .

Decreasing on $(7, \infty)$ since $f' (x) < 0$

Step 8

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

Decreasing on: $(- \infty, 7), (7, \infty)$

Step 9