Calculus Examples

$f (x) = x + \sqrt{x}$

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 $x + \sqrt{x}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [\sqrt{x}]$ .

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

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 $n = 1$ .

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

Step 1.1.1.2

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

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Step 1.1.1.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

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

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 $n = \frac{1}{2}$ .

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

Step 1.1.1.2.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 1.1.1.2.4

Combine $- 1$ and $\frac{2}{2}$ .

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

Step 1.1.1.2.5

Combine the numerators over the common denominator.

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

Step 1.1.1.2.6

Simplify the numerator.

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Step 1.1.1.2.6.1

Multiply $- 1$ by $2$ .

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

Step 1.1.1.2.6.2

Subtract $2$ from $1$ .

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

Step 1.1.1.2.7

Move the negative in front of the fraction.

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

Step 1.1.1.3

Simplify.

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Step 1.1.1.3.1

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

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

Step 1.1.1.3.2

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

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

Differentiate.

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Step 1.1.2.1.1

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

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

Step 1.1.2.1.2

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

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

Step 1.1.2.2

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

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Step 1.1.2.2.1

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

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

Step 1.1.2.2.2

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

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

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 $g (x) = x^{\frac{1}{2}}$ .

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Step 1.1.2.2.3.1

To apply the Chain Rule, set $u$ as $x^{\frac{1}{2}}$ .

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

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 $n = - 1$ .

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

Step 1.1.2.2.3.3

Replace all occurrences of $u$ with $x^{\frac{1}{2}}$ .

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

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 $n = \frac{1}{2}$ .

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

Step 1.1.2.2.5

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

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Step 1.1.2.2.5.1

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

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

Step 1.1.2.2.5.2

Cancel the common factor of $2$ .

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Step 1.1.2.2.5.2.1

Factor $2$ out of $- 2$ .

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

Step 1.1.2.2.5.2.2

Cancel the common factor.

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

Step 1.1.2.2.5.2.3

Rewrite the expression.

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

Step 1.1.2.2.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 1.1.2.2.7

Combine $- 1$ and $\frac{2}{2}$ .

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

Step 1.1.2.2.8

Combine the numerators over the common denominator.

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

Step 1.1.2.2.9

Simplify the numerator.

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Step 1.1.2.2.9.1

Multiply $- 1$ by $2$ .

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

Step 1.1.2.2.9.2

Subtract $2$ from $1$ .

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

Step 1.1.2.2.10

Move the negative in front of the fraction.

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

Step 1.1.2.2.11

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

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

Step 1.1.2.2.12

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

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

Step 1.1.2.2.13

Multiply $x^{- \frac{1}{2}}$ by $x^{- 1}$ by adding the exponents.

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Step 1.1.2.2.13.1

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

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

Step 1.1.2.2.13.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 1.1.2.2.13.3

Combine $- 1$ and $\frac{2}{2}$ .

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

Step 1.1.2.2.13.4

Combine the numerators over the common denominator.

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

Step 1.1.2.2.13.5

Simplify the numerator.

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Step 1.1.2.2.13.5.1

Multiply $- 1$ by $2$ .

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

Step 1.1.2.2.13.5.2

Subtract $2$ from $- 1$ .

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

Step 1.1.2.2.13.6

Move the negative in front of the fraction.

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

Step 1.1.2.2.14

Move $x^{- \frac{3}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.1.2.2.15

Multiply $\frac{1}{2}$ by $\frac{1}{2 x^{\frac{3}{2}}}$ .

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

Step 1.1.2.2.16

Multiply $2$ by $2$ .

$f'' (x) = 0 - \frac{1}{4 x^{\frac{3}{2}}}$

Step 1.1.2.3

Subtract $\frac{1}{4 x^{\frac{3}{2}}}$ from $0$ .

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

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $- \frac{1}{4 x^{\frac{3}{2}}}$ .

$- \frac{1}{4 x^{\frac{3}{2}}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $- \frac{1}{4 x^{\frac{3}{2}}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$- \frac{1}{4 x^{\frac{3}{2}}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$1 = 0$

Step 1.2.3

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

No solution

Step 2

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

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Step 2.1

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

Step 2.2

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

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Step 3

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

$[0, \infty)$

Step 4

Substitute any number from the interval $[0, \infty)$ 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.

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

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

Rewrite $4$ as $2^{2}$ .

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

Step 4.2.1.2

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

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

Step 4.2.1.3

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 4.2.1.4

Combine $2$ and $\frac{2}{2}$ .

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

Step 4.2.1.5

Combine the numerators over the common denominator.

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

Step 4.2.1.6

Simplify the numerator.

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Step 4.2.1.6.1

Multiply $2$ by $2$ .

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

Step 4.2.1.6.2

Add $4$ and $3$ .

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

Step 4.2.2

The final answer is $- \frac{1}{2^{\frac{7}{2}}}$ .

$- \frac{1}{2^{\frac{7}{2}}}$

Step 4.3

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

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

Step 5