Calculus Examples

$f (x) = \frac{3}{x - 5}$

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

Differentiate using the Constant Multiple Rule.

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Step 1.1.1.1.1

Since $3$ is constant with respect to $x$ , the derivative of $\frac{3}{x - 5}$ with respect to $x$ is $3 \frac{d}{d x} [\frac{1}{x - 5}]$ .

$f' (x) = 3 \frac{d}{d x} (\frac{1}{x - 5})$

Step 1.1.1.1.2

Rewrite $\frac{1}{x - 5}$ as ${(x - 5)}^{- 1}$ .

$f' (x) = 3 \frac{d}{d x} ({(x - 5)}^{- 1})$

Step 1.1.1.2

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 $g (x) = x - 5$ .

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Step 1.1.1.2.1

To apply the Chain Rule, set $u$ as $x - 5$ .

$f' (x) = 3 (\frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x - 5))$

Step 1.1.1.2.2

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

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

Step 1.1.1.2.3

Replace all occurrences of $u$ with $x - 5$ .

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

Step 1.1.1.3

Differentiate.

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Step 1.1.1.3.1

Multiply $- 1$ by $3$ .

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

Step 1.1.1.3.2

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

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

Step 1.1.1.3.3

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

$f' (x) = - 3 {(x - 5)}^{- 2} (1 + \frac{d}{d x} (- 5))$

Step 1.1.1.3.4

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$f' (x) = - 3 {(x - 5)}^{- 2} (1 + 0)$

Step 1.1.1.3.5

Simplify the expression.

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Step 1.1.1.3.5.1

Add $1$ and $0$ .

$f' (x) = - 3 {(x - 5)}^{- 2} \cdot 1$

Step 1.1.1.3.5.2

Multiply $- 3$ by $1$ .

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

Step 1.1.1.4

Simplify.

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Step 1.1.1.4.1

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

$f' (x) = - 3 \frac{1}{{(x - 5)}^{2}}$

Step 1.1.1.4.2

Combine terms.

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Step 1.1.1.4.2.1

Combine $- 3$ and $\frac{1}{{(x - 5)}^{2}}$ .

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

Step 1.1.1.4.2.2

Move the negative in front of the fraction.

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

Differentiate using the Constant Multiple Rule.

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Step 1.1.2.1.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- \frac{3}{{(x - 5)}^{2}}$ with respect to $x$ is $- 3 \frac{d}{d x} [\frac{1}{{(x - 5)}^{2}}]$ .

$f'' (x) = - 3 \frac{d}{d x} \frac{1}{{(x - 5)}^{2}}$

Step 1.1.2.1.2

Apply basic rules of exponents.

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Step 1.1.2.1.2.1

Rewrite $\frac{1}{{(x - 5)}^{2}}$ as ${({(x - 5)}^{2})}^{- 1}$ .

$f'' (x) = - 3 \frac{d}{d x} {({(x - 5)}^{2})}^{- 1}$

Step 1.1.2.1.2.2

Multiply the exponents in ${({(x - 5)}^{2})}^{- 1}$ .

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Step 1.1.2.1.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f'' (x) = - 3 \frac{d}{d x} {(x - 5)}^{2 \cdot - 1}$

Step 1.1.2.1.2.2.2

Multiply $2$ by $- 1$ .

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

Step 1.1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 2}$ and $g (x) = x - 5$ .

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Step 1.1.2.2.1

To apply the Chain Rule, set $u$ as $x - 5$ .

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

Step 1.1.2.2.2

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

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

Step 1.1.2.2.3

Replace all occurrences of $u$ with $x - 5$ .

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

Step 1.1.2.3

Differentiate.

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Step 1.1.2.3.1

Multiply $- 2$ by $- 3$ .

$f'' (x) = 6 ({(x - 5)}^{- 3} \frac{d}{d x} (x - 5))$

Step 1.1.2.3.2

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

$f'' (x) = 6 {(x - 5)}^{- 3} (\frac{d}{d x} (x) + \frac{d}{d x} (- 5))$

Step 1.1.2.3.3

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

$f'' (x) = 6 {(x - 5)}^{- 3} (1 + \frac{d}{d x} (- 5))$

Step 1.1.2.3.4

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$f'' (x) = 6 {(x - 5)}^{- 3} (1 + 0)$

Step 1.1.2.3.5

Simplify the expression.

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Step 1.1.2.3.5.1

Add $1$ and $0$ .

$f'' (x) = 6 {(x - 5)}^{- 3} \cdot 1$

Step 1.1.2.3.5.2

Multiply $6$ by $1$ .

$f'' (x) = 6 {(x - 5)}^{- 3}$

Step 1.1.2.4

Simplify.

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Step 1.1.2.4.1

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

$f'' (x) = 6 (\frac{1}{{(x - 5)}^{3}})$

Step 1.1.2.4.2

Combine $6$ and $\frac{1}{{(x - 5)}^{3}}$ .

$f'' (x) = \frac{6}{{(x - 5)}^{3}}$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{6}{{(x - 5)}^{3}}$ .

$\frac{6}{{(x - 5)}^{3}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $\frac{6}{{(x - 5)}^{3}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$\frac{6}{{(x - 5)}^{3}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$6 = 0$

Step 1.2.3

Since $6 \neq 0$ , there are no solutions.

No solution

Step 2

Find the domain of $f (x) = \frac{3}{x - 5}$ .

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Step 2.1

Set the denominator in $\frac{3}{x - 5}$ equal to $0$ to find where the expression is undefined.

$x - 5 = 0$

Step 2.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, 5) \cup (5, \infty)$

Set-Builder Notation:

${x | x \neq 5}$

Interval Notation:

$(- \infty, 5) \cup (5, \infty)$

Set-Builder Notation:

${x | x \neq 5}$

Step 3

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, 5) \cup (5, \infty)$

Step 4

Substitute any number from the interval $(- \infty, 5)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

Replace the variable $x$ with $0$ in the expression.

$f'' (0) = \frac{6}{{((0) - 5)}^{3}}$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify the denominator.

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Step 4.2.1.1

Subtract $5$ from $0$ .

$f'' (0) = \frac{6}{{(- 5)}^{3}}$

Step 4.2.1.2

Raise $- 5$ to the power of $3$ .

$f'' (0) = \frac{6}{- 125}$

Step 4.2.2

Move the negative in front of the fraction.

$f'' (0) = - \frac{6}{125}$

Step 4.2.3

The final answer is $- \frac{6}{125}$ .

$- \frac{6}{125}$

Step 4.3

The graph is concave down on the interval $(- \infty, 5)$ because $f'' (0)$ is negative.

Concave down on $(- \infty, 5)$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(5, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

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

$f'' (8) = \frac{6}{{((8) - 5)}^{3}}$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify the denominator.

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Step 5.2.1.1

Subtract $5$ from $8$ .

$f'' (8) = \frac{6}{3^{3}}$

Step 5.2.1.2

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

$f'' (8) = \frac{6}{27}$

Step 5.2.2

Cancel the common factor of $6$ and $27$ .

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Step 5.2.2.1

Factor $3$ out of $6$ .

$f'' (8) = \frac{3 (2)}{27}$

Step 5.2.2.2

Cancel the common factors.

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Step 5.2.2.2.1

Factor $3$ out of $27$ .

$f'' (8) = \frac{3 \cdot 2}{3 \cdot 9}$

Step 5.2.2.2.2

Cancel the common factor.

$f'' (8) = \frac{3 \cdot 2}{3 \cdot 9}$

Step 5.2.2.2.3

Rewrite the expression.

$f'' (8) = \frac{2}{9}$

Step 5.2.3

The final answer is $\frac{2}{9}$ .

$\frac{2}{9}$

Step 5.3

The graph is concave up on the interval $(5, \infty)$ because $f'' (8)$ is positive.

Concave up on $(5, \infty)$ since $f'' (x)$ is positive

Step 6

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

Concave down on $(- \infty, 5)$ since $f'' (x)$ is negative

Concave up on $(5, \infty)$ since $f'' (x)$ is positive

Step 7