Calculus Examples

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

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 $x - 3 x^{\frac{1}{3}}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3 x^{\frac{1}{3}}]$ .

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

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 $n = 1$ .

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

Step 1.1.1.2

Evaluate $\frac{d}{d x} [- 3 x^{\frac{1}{3}}]$ .

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Step 1.1.1.2.1

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

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

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 $n = \frac{1}{3}$ .

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

Step 1.1.1.2.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 1.1.1.2.4

Combine $- 1$ and $\frac{3}{3}$ .

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

Step 1.1.1.2.5

Combine the numerators over the common denominator.

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

Step 1.1.1.2.6

Simplify the numerator.

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Step 1.1.1.2.6.1

Multiply $- 1$ by $3$ .

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

Step 1.1.1.2.6.2

Subtract $3$ from $1$ .

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

Step 1.1.1.2.7

Move the negative in front of the fraction.

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

Step 1.1.1.2.8

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

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

Step 1.1.1.2.9

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

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

Step 1.1.1.2.10

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

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

Step 1.1.1.2.11

Factor $3$ out of $- 3$ .

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

Step 1.1.1.2.12

Cancel the common factors.

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Step 1.1.1.2.12.1

Factor $3$ out of $3 x^{\frac{2}{3}}$ .

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

Step 1.1.1.2.12.2

Cancel the common factor.

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

Step 1.1.1.2.12.3

Rewrite the expression.

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

Step 1.1.1.2.13

Move the negative in front of the fraction.

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

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 $1 - \frac{1}{x^{\frac{2}{3}}}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- \frac{1}{x^{\frac{2}{3}}}]$ .

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

Step 1.1.2.1.2

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

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

Step 1.1.2.2

Evaluate $\frac{d}{d x} [- \frac{1}{x^{\frac{2}{3}}}]$ .

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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 $g (x) = \frac{1}{x^{\frac{2}{3}}}$ .

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

Step 1.1.2.2.2

Rewrite $\frac{1}{x^{\frac{2}{3}}}$ as ${(x^{\frac{2}{3}})}^{- 1}$ .

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

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 $g (x) = x^{\frac{2}{3}}$ .

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Step 1.1.2.2.3.1

To apply the Chain Rule, set $u$ as $x^{\frac{2}{3}}$ .

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

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 $n = - 1$ .

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

Step 1.1.2.2.3.3

Replace all occurrences of $u$ with $x^{\frac{2}{3}}$ .

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

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 $n = \frac{2}{3}$ .

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

Step 1.1.2.2.5

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

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

Step 1.1.2.2.6

Multiply the exponents in ${(x^{\frac{2}{3}})}^{- 2}$ .

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Step 1.1.2.2.6.1

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

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

Step 1.1.2.2.6.2

Multiply $\frac{2}{3} \cdot - 2$ .

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Step 1.1.2.2.6.2.1

Combine $\frac{2}{3}$ and $- 2$ .

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

Step 1.1.2.2.6.2.2

Multiply $2$ by $- 2$ .

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

Step 1.1.2.2.6.3

Move the negative in front of the fraction.

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

Step 1.1.2.2.7

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 1.1.2.2.8

Combine $- 1$ and $\frac{3}{3}$ .

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

Step 1.1.2.2.9

Combine the numerators over the common denominator.

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

Step 1.1.2.2.10

Simplify the numerator.

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Step 1.1.2.2.10.1

Multiply $- 1$ by $3$ .

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

Step 1.1.2.2.10.2

Subtract $3$ from $2$ .

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

Step 1.1.2.2.11

Move the negative in front of the fraction.

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

Step 1.1.2.2.12

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

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

Step 1.1.2.2.13

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

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

Step 1.1.2.2.14

Multiply $x^{- \frac{1}{3}}$ by $x^{- \frac{4}{3}}$ by adding the exponents.

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Step 1.1.2.2.14.1

Move $x^{- \frac{4}{3}}$ .

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

Step 1.1.2.2.14.2

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

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

Step 1.1.2.2.14.3

Combine the numerators over the common denominator.

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

Step 1.1.2.2.14.4

Subtract $1$ from $- 4$ .

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

Step 1.1.2.2.14.5

Move the negative in front of the fraction.

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

Step 1.1.2.2.15

Move $x^{- \frac{5}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.1.2.2.16

Multiply $- 1$ by $- 1$ .

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

Step 1.1.2.2.17

Multiply $\frac{2}{3 x^{\frac{5}{3}}}$ by $1$ .

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

Step 1.1.2.2.18

Multiply $\frac{1}{x^{\frac{2}{3}}}$ by $0$ .

$f'' (x) = 0 + \frac{2}{3 x^{\frac{5}{3}}} + 0$

Step 1.1.2.2.19

Add $\frac{2}{3 x^{\frac{5}{3}}}$ and $0$ .

$f'' (x) = 0 + \frac{2}{3 x^{\frac{5}{3}}}$

Step 1.1.2.3

Add $0$ and $\frac{2}{3 x^{\frac{5}{3}}}$ .

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

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{2}{3 x^{\frac{5}{3}}}$ .

$\frac{2}{3 x^{\frac{5}{3}}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $\frac{2}{3 x^{\frac{5}{3}}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$\frac{2}{3 x^{\frac{5}{3}}} = 0$

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

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Step 2.1

Convert expressions with fractional exponents to radicals.

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Step 2.1.1

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$x - 3 \sqrt[3]{x^{1}}$

Step 2.1.2

Anything raised to $1$ is the base itself.

$x - 3 \sqrt[3]{x}$

Step 2.2

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3

The graph is concave up because the second derivative is positive.

The graph is concave up

Step 4