Calculus Examples

$\frac{1}{x - 4}$

Write $\frac{1}{x - 4}$ as a function.

$f (x) = \frac{1}{x - 4}$

Find the first derivative.

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Find the first derivative.

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Rewrite $\frac{1}{x - 4}$ as ${(x - 4)}^{- 1}$ .

$f' (x) = \frac{d}{d x} ({(x - 4)}^{- 1})$

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 $g (x) = x - 4$ .

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To apply the Chain Rule, set $u$ as $x - 4$ .

$f' (x) = \frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x - 4)$

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

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

Replace all occurrences of $u$ with $x - 4$ .

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

Differentiate.

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

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

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

$f' (x) = - {(x - 4)}^{- 2} (1 + \frac{d}{d x} (- 4))$

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

$f' (x) = - {(x - 4)}^{- 2} (1 + 0)$

Simplify the expression.

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Add $1$ and $0$ .

$f' (x) = - {(x - 4)}^{- 2} \cdot 1$

Multiply $- 1$ by $1$ .

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

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

$f' (x) = - \frac{1}{{(x - 4)}^{2}}$

The first derivative of $f (x)$ with respect to $x$ is $- \frac{1}{{(x - 4)}^{2}}$ .

$- \frac{1}{{(x - 4)}^{2}}$

Set the first derivative equal to $0$ then solve the equation $- \frac{1}{{(x - 4)}^{2}} = 0$ .

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Set the first derivative equal to $0$ .

$- \frac{1}{{(x - 4)}^{2}} = 0$

Set the numerator equal to zero.

$1 = 0$

Since $1 \neq 0$ , there are no solutions.

No solution

Find where the derivative is undefined.

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Set the denominator in $\frac{1}{{(x - 4)}^{2}}$ equal to $0$ to find where the expression is undefined.

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

Solve for $x$ .

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Set the $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

After finding the point that makes the derivative $f' (x) = - \frac{1}{{(x - 4)}^{2}}$ equal to $0$ or undefined, the interval to check where $f (x) = \frac{1}{x - 4}$ is increasing and where it is decreasing is $(- \infty, 4) \cup (4, \infty)$ .

$(- \infty, 4) \cup (4, \infty)$

Substitute a value from the interval $(- \infty, 4)$ 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.

$f' (3) = - \frac{1}{{((3) - 4)}^{2}}$

Simplify the result.

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Simplify the denominator.

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Subtract $4$ from $3$ .

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

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

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

Reduce the expression by cancelling the common factors.

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Cancel the common factor of $1$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $1$ .

$f' (3) = - 1$

The final answer is $- 1$ .

$- 1$

At $x = 3$ the derivative is $- 1$ . Since this is negative, the function is decreasing on $(- \infty, 4)$ .

Decreasing on $(- \infty, 4)$ since $f' (x) < 0$

Substitute a value from the interval $(4, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Replace the variable $x$ with $5$ in the expression.

$f' (5) = - \frac{1}{{((5) - 4)}^{2}}$

Simplify the result.

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Simplify the denominator.

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Subtract $4$ from $5$ .

$f' (5) = - \frac{1}{1^{2}}$

One to any power is one.

$f' (5) = - \frac{1}{1}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor of $1$ .

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Cancel the common factor.

$f' (5) = - \frac{1}{1}$

Rewrite the expression.

$f' (5) = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$f' (5) = - 1$

The final answer is $- 1$ .

$- 1$

At $x = 5$ the derivative is $- 1$ . Since this is negative, the function is decreasing on $(4, \infty)$ .

Decreasing on $(4, \infty)$ since $f' (x) < 0$

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

Decreasing on: $(- \infty, 4), (4, \infty)$