Calculus Examples

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

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 Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x + 1$ and $g (x) = x - 1$ .

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

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

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

Step 1.1.1.2.2

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

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

Step 1.1.1.2.3

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f' (x) = \frac{(x - 1) (1 + 0) - (x + 1) \frac{d}{d x} (x - 1)}{{(x - 1)}^{2}}$

Step 1.1.1.2.4

Simplify the expression.

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Step 1.1.1.2.4.1

Add $1$ and $0$ .

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

Step 1.1.1.2.4.2

Multiply $x - 1$ by $1$ .

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

Step 1.1.1.2.5

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

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

Step 1.1.1.2.6

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

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

Step 1.1.1.2.7

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

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

Step 1.1.1.2.8

Simplify the expression.

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Step 1.1.1.2.8.1

Add $1$ and $0$ .

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

Step 1.1.1.2.8.2

Multiply $- 1$ by $1$ .

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

Step 1.1.1.3

Simplify.

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Step 1.1.1.3.1

Apply the distributive property.

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

Step 1.1.1.3.2

Simplify the numerator.

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Step 1.1.1.3.2.1

Combine the opposite terms in $x - 1 - x - 1 \cdot 1$ .

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Step 1.1.1.3.2.1.1

Subtract $x$ from $x$ .

$f' (x) = \frac{0 - 1 - 1 \cdot 1}{{(x - 1)}^{2}}$

Step 1.1.1.3.2.1.2

Subtract $1$ from $0$ .

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

Step 1.1.1.3.2.2

Multiply $- 1$ by $1$ .

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

Step 1.1.1.3.2.3

Subtract $1$ from $- 1$ .

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

Step 1.1.1.3.3

Move the negative in front of the fraction.

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

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

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

Step 1.1.2.1.2

Apply basic rules of exponents.

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Step 1.1.2.1.2.1

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

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

Step 1.1.2.1.2.2

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

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Step 1.1.2.1.2.2.1

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

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

Step 1.1.2.1.2.2.2

Multiply $2$ by $- 1$ .

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

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 $g (x) = x - 1$ .

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Step 1.1.2.2.1

To apply the Chain Rule, set $u$ as $x - 1$ .

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

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 $n = - 2$ .

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

Step 1.1.2.2.3

Replace all occurrences of $u$ with $x - 1$ .

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

Step 1.1.2.3

Differentiate.

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Step 1.1.2.3.1

Multiply $- 2$ by $- 2$ .

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

Step 1.1.2.3.2

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

$f'' (x) = 4 {(x - 1)}^{- 3} (\frac{d}{d x} (x) + \frac{d}{d x} (- 1))$

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 $n = 1$ .

$f'' (x) = 4 {(x - 1)}^{- 3} (1 + \frac{d}{d x} (- 1))$

Step 1.1.2.3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f'' (x) = 4 {(x - 1)}^{- 3} (1 + 0)$

Step 1.1.2.3.5

Simplify the expression.

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Step 1.1.2.3.5.1

Add $1$ and $0$ .

$f'' (x) = 4 {(x - 1)}^{- 3} \cdot 1$

Step 1.1.2.3.5.2

Multiply $4$ by $1$ .

$f'' (x) = 4 {(x - 1)}^{- 3}$

Step 1.1.2.4

Simplify.

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Step 1.1.2.4.1

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

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

Step 1.1.2.4.2

Combine $4$ and $\frac{1}{{(x - 1)}^{3}}$ .

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

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{4}{{(x - 1)}^{3}}$ .

$\frac{4}{{(x - 1)}^{3}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $\frac{4}{{(x - 1)}^{3}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$\frac{4}{{(x - 1)}^{3}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$4 = 0$

Step 1.2.3

Since $4 \neq 0$ , there are no solutions.

No solution

Step 2

Find the domain of $f (x) = \frac{x + 1}{x - 1}$ .

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Step 2.1

Set the denominator in $\frac{x + 1}{x - 1}$ equal to $0$ to find where the expression is undefined.

$x - 1 = 0$

Step 2.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.3

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

Interval Notation:

$(- \infty, 1) \cup (1, \infty)$

Set-Builder Notation:

${x | x \neq 1}$

Interval Notation:

$(- \infty, 1) \cup (1, \infty)$

Set-Builder Notation:

${x | x \neq 1}$

Step 3

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

$(- \infty, 1) \cup (1, \infty)$

Step 4

Substitute any number from the interval $(- \infty, 1)$ 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.

$f'' (0) = \frac{4}{{((0) - 1)}^{3}}$

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 $1$ from $0$ .

$f'' (0) = \frac{4}{{(- 1)}^{3}}$

Step 4.2.1.2

Raise $- 1$ to the power of $3$ .

$f'' (0) = \frac{4}{- 1}$

Step 4.2.2

Divide $4$ by $- 1$ .

$f'' (0) = - 4$

Step 4.2.3

The final answer is $- 4$ .

$- 4$

Step 4.3

The graph is concave down on the interval $(- \infty, 1)$ because $f'' (0)$ is negative.

Concave down on $(- \infty, 1)$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(1, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

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

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

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 $1$ from $4$ .

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

Step 5.2.1.2

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

$f'' (4) = \frac{4}{27}$

Step 5.2.2

The final answer is $\frac{4}{27}$ .

$\frac{4}{27}$

Step 5.3

The graph is concave up on the interval $(1, \infty)$ because $f'' (4)$ is positive.

Concave up on $(1, \infty)$ 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 $(- \infty, 1)$ since $f'' (x)$ is negative

Concave up on $(1, \infty)$ since $f'' (x)$ is positive

Step 7