Calculus Examples

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

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.

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Step 1.1.1.1.1

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

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

Step 1.1.1.1.2

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

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

Step 1.1.1.2

Evaluate $\frac{d}{d x} [\frac{1}{x}]$ .

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Step 1.1.1.2.1

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

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

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 $n = - 1$ .

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

Step 1.1.1.3

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

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

Step 1.1.1.4

Reorder terms.

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

By the Sum Rule, the derivative of $- \frac{1}{x^{2}} + 1$ with respect to $x$ is $\frac{d}{d x} [- \frac{1}{x^{2}}] + \frac{d}{d x} [1]$ .

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

Step 1.1.2.2

Evaluate $\frac{d}{d x} [- \frac{1}{x^{2}}]$ .

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Step 1.1.2.2.1

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

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

Step 1.1.2.2.2

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

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

Step 1.1.2.2.3

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 $g (x) = x^{2}$ .

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Step 1.1.2.2.3.1

To apply the Chain Rule, set $u$ as $x^{2}$ .

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

Step 1.1.2.2.3.2

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

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

Step 1.1.2.2.3.3

Replace all occurrences of $u$ with $x^{2}$ .

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

Step 1.1.2.2.4

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

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

Step 1.1.2.2.5

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

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

Step 1.1.2.2.6

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

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Step 1.1.2.2.6.1

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

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

Step 1.1.2.2.6.2

Multiply $2$ by $- 2$ .

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

Step 1.1.2.2.7

Multiply $2$ by $- 1$ .

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

Step 1.1.2.2.8

Raise $x$ to the power of $1$ .

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

Step 1.1.2.2.9

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 1.1.2.2.10

Subtract $4$ from $1$ .

$f'' (x) = - (- 2 x^{- 3}) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} (1)$

Step 1.1.2.2.11

Multiply $- 2$ by $- 1$ .

$f'' (x) = 2 x^{- 3} + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} (1)$

Step 1.1.2.2.12

Multiply $\frac{1}{x^{2}}$ by $0$ .

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

Step 1.1.2.2.13

Add $2 x^{- 3}$ and $0$ .

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

Step 1.1.2.3

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

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

Step 1.1.2.4

Simplify.

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Step 1.1.2.4.1

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

$f'' (x) = 2 (\frac{1}{x^{3}}) + 0$

Step 1.1.2.4.2

Combine terms.

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Step 1.1.2.4.2.1

Combine $2$ and $\frac{1}{x^{3}}$ .

$f'' (x) = \frac{2}{x^{3}} + 0$

Step 1.1.2.4.2.2

Add $\frac{2}{x^{3}}$ and $0$ .

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

Step 1.1.3

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

$\frac{2}{x^{3}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $\frac{2}{x^{3}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$\frac{2}{x^{3}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$2 = 0$

Step 1.2.3

Since $2 \neq 0$ , there are no solutions.

No solution

Step 2

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

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Step 2.1

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

$x = 0$

Step 2.2

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

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0}$

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0}$

Step 3

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

$(- \infty, 0) \cup (0, \infty)$

Step 4

Substitute any number from the interval $(- \infty, 0)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

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

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

Step 4.2

Simplify the result.

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Step 4.2.1

Cancel the common factor of $2$ and ${(- 2)}^{3}$ .

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Step 4.2.1.1

Rewrite $2$ as $- 1 (- 2)$ .

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

Step 4.2.1.2

Factor $- 2$ out of $- 1 (- 2)$ .

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

Step 4.2.1.3

Cancel the common factors.

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Step 4.2.1.3.1

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

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

Step 4.2.1.3.2

Cancel the common factor.

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

Step 4.2.1.3.3

Rewrite the expression.

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

Step 4.2.2

Simplify the expression.

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Step 4.2.2.1

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

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

Step 4.2.2.2

Move the negative in front of the fraction.

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

Step 4.2.3

The final answer is $- \frac{1}{4}$ .

$- \frac{1}{4}$

Step 4.3

The graph is concave down on the interval $(- \infty, 0)$ because $f'' (- 2)$ is negative.

Concave down on $(- \infty, 0)$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(0, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

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

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

Step 5.2

Simplify the result.

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Step 5.2.1

Cancel the common factor of $2$ and ${(2)}^{3}$ .

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Step 5.2.1.1

Factor $2$ out of $2$ .

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

Step 5.2.1.2

Cancel the common factors.

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Step 5.2.1.2.1

Factor $2$ out of ${(2)}^{3}$ .

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

Step 5.2.1.2.2

Cancel the common factor.

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

Step 5.2.1.2.3

Rewrite the expression.

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

Step 5.2.2

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

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

Step 5.2.3

The final answer is $\frac{1}{4}$ .

$\frac{1}{4}$

Step 5.3

The graph is concave up on the interval $(0, \infty)$ because $f'' (2)$ is positive.

Concave up on $(0, \infty)$ 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 $(- \infty, 0)$ since $f'' (x)$ is negative

Concave up on $(0, \infty)$ since $f'' (x)$ is positive

Step 7