Calculus Examples

$h (x) = x^{5} + 5 x^{4}$

Step 1

Find the second derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate.

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Step 1.1.1.1

By the Sum Rule, the derivative of $x^{5} + 5 x^{4}$ with respect to $x$ is $\frac{d}{d x} [x^{5}] + \frac{d}{d x} [5 x^{4}]$ .

$\frac{d}{d x} [x^{5}] + \frac{d}{d x} [5 x^{4}]$

Step 1.1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$5 x^{4} + \frac{d}{d x} [5 x^{4}]$

Step 1.1.2

Evaluate $\frac{d}{d x} [5 x^{4}]$ .

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Step 1.1.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{4}$ with respect to $x$ is $5 \frac{d}{d x} [x^{4}]$ .

$5 x^{4} + 5 \frac{d}{d x} [x^{4}]$

Step 1.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$5 x^{4} + 5 (4 x^{3})$

Step 1.1.2.3

Multiply $4$ by $5$ .

$f' (x) = 5 x^{4} + 20 x^{3}$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $5 x^{4} + 20 x^{3}$ with respect to $x$ is $\frac{d}{d x} [5 x^{4}] + \frac{d}{d x} [20 x^{3}]$ .

$\frac{d}{d x} [5 x^{4}] + \frac{d}{d x} [20 x^{3}]$

Step 1.2.2

Evaluate $\frac{d}{d x} [5 x^{4}]$ .

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Step 1.2.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{4}$ with respect to $x$ is $5 \frac{d}{d x} [x^{4}]$ .

$5 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [20 x^{3}]$

Step 1.2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$5 (4 x^{3}) + \frac{d}{d x} [20 x^{3}]$

Step 1.2.2.3

Multiply $4$ by $5$ .

$20 x^{3} + \frac{d}{d x} [20 x^{3}]$

Step 1.2.3

Evaluate $\frac{d}{d x} [20 x^{3}]$ .

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Step 1.2.3.1

Since $20$ is constant with respect to $x$ , the derivative of $20 x^{3}$ with respect to $x$ is $20 \frac{d}{d x} [x^{3}]$ .

$20 x^{3} + 20 \frac{d}{d x} [x^{3}]$

Step 1.2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$20 x^{3} + 20 (3 x^{2})$

Step 1.2.3.3

Multiply $3$ by $20$ .

$f'' (x) = 20 x^{3} + 60 x^{2}$

Step 1.3

The second derivative of $h (x)$ with respect to $x$ is $20 x^{3} + 60 x^{2}$ .

$20 x^{3} + 60 x^{2}$

Step 2

Set the second derivative equal to $0$ then solve the equation $20 x^{3} + 60 x^{2} = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$20 x^{3} + 60 x^{2} = 0$

Step 2.2

Factor $20 x^{2}$ out of $20 x^{3} + 60 x^{2}$ .

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Step 2.2.1

Factor $20 x^{2}$ out of $20 x^{3}$ .

$20 x^{2} (x) + 60 x^{2} = 0$

Step 2.2.2

Factor $20 x^{2}$ out of $60 x^{2}$ .

$20 x^{2} (x) + 20 x^{2} (3) = 0$

Step 2.2.3

Factor $20 x^{2}$ out of $20 x^{2} (x) + 20 x^{2} (3)$ .

$20 x^{2} (x + 3) = 0$

Step 2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x^{2} = 0$

$x + 3 = 0$

Step 2.4

Set $x^{2}$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Step 2.4.2

Solve $x^{2} = 0$ for $x$ .

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Step 2.4.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Step 2.4.2.2

Simplify $\pm \sqrt{0}$ .

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Step 2.4.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 2.4.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Step 2.4.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 2.5

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.5.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.6

The final solution is all the values that make $20 x^{2} (x + 3) = 0$ true.

$x = 0, - 3$

Step 3

Find the points where the second derivative is $0$ .

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Step 3.1

Substitute $0$ in $h (x) = x^{5} + 5 x^{4}$ to find the value of $y$ .

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Step 3.1.1

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

$h (0) = {(0)}^{5} + 5 {(0)}^{4}$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

Raising $0$ to any positive power yields $0$ .

$h (0) = 0 + 5 {(0)}^{4}$

Step 3.1.2.1.2

Raising $0$ to any positive power yields $0$ .

$h (0) = 0 + 5 \cdot 0$

Step 3.1.2.1.3

Multiply $5$ by $0$ .

$h (0) = 0 + 0$

Step 3.1.2.2

Add $0$ and $0$ .

$h (0) = 0$

Step 3.1.2.3

The final answer is $0$ .

$0$

Step 3.2

The point found by substituting $0$ in $h (x) = x^{5} + 5 x^{4}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Step 3.3

Substitute $- 3$ in $h (x) = x^{5} + 5 x^{4}$ to find the value of $y$ .

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Step 3.3.1

Replace the variable $x$ with $- 3$ in the expression.

$h (- 3) = {(- 3)}^{5} + 5 {(- 3)}^{4}$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

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

$h (- 3) = - 243 + 5 {(- 3)}^{4}$

Step 3.3.2.1.2

Raise $- 3$ to the power of $4$ .

$h (- 3) = - 243 + 5 \cdot 81$

Step 3.3.2.1.3

Multiply $5$ by $81$ .

$h (- 3) = - 243 + 405$

Step 3.3.2.2

Add $- 243$ and $405$ .

$h (- 3) = 162$

Step 3.3.2.3

The final answer is $162$ .

$162$

Step 3.4

The point found by substituting $- 3$ in $h (x) = x^{5} + 5 x^{4}$ is $(- 3, 162)$ . This point can be an inflection point.

$(- 3, 162)$

Step 3.5

Determine the points that could be inflection points.

$(0, 0), (- 3, 162)$

Step 4

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, - 3) \cup (- 3, 0) \cup (0, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 3)$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 3.1$ in the expression.

$h'' (- 3.1) = 20 {(- 3.1)}^{3} + 60 {(- 3.1)}^{2}$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Raise $- 3.1$ to the power of $3$ .

$h'' (- 3.1) = 20 \cdot - 29.791 + 60 {(- 3.1)}^{2}$

Step 5.2.1.2

Multiply $20$ by $- 29.791$ .

$h'' (- 3.1) = - 595.82 + 60 {(- 3.1)}^{2}$

Step 5.2.1.3

Raise $- 3.1$ to the power of $2$ .

$h'' (- 3.1) = - 595.82 + 60 \cdot 9.61$

Step 5.2.1.4

Multiply $60$ by $9.61$ .

$h'' (- 3.1) = - 595.82 + 576.6$

Step 5.2.2

Add $- 595.82$ and $576.6$ .

$h'' (- 3.1) = - 19.22$

Step 5.2.3

The final answer is $- 19.22$ .

$- 19.22$

Step 5.3

At $- 3.1$ , the second derivative is $- 19.22$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, - 3)$

Decreasing on $(- \infty, - 3)$ since $h'' (x) < 0$

Step 6

Substitute a value from the interval $(- 3, 0)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- \frac{3}{2}$ in the expression.

$h'' (- \frac{3}{2}) = 20 {(- \frac{3}{2})}^{3} + 60 {(- \frac{3}{2})}^{2}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 6.2.1.1.1

Apply the product rule to $- \frac{3}{2}$ .

$h'' (- \frac{3}{2}) = 20 ({(- 1)}^{3} {(\frac{3}{2})}^{3}) + 60 {(- \frac{3}{2})}^{2}$

Step 6.2.1.1.2

Apply the product rule to $\frac{3}{2}$ .

$h'' (- \frac{3}{2}) = 20 ({(- 1)}^{3} (\frac{3^{3}}{2^{3}})) + 60 {(- \frac{3}{2})}^{2}$

Step 6.2.1.2

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

$h'' (- \frac{3}{2}) = 20 (- \frac{3^{3}}{2^{3}}) + 60 {(- \frac{3}{2})}^{2}$

Step 6.2.1.3

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

$h'' (- \frac{3}{2}) = 20 (- \frac{27}{2^{3}}) + 60 {(- \frac{3}{2})}^{2}$

Step 6.2.1.4

Raise $2$ to the power of $3$ .

$h'' (- \frac{3}{2}) = 20 (- \frac{27}{8}) + 60 {(- \frac{3}{2})}^{2}$

Step 6.2.1.5

Cancel the common factor of $4$ .

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Step 6.2.1.5.1

Move the leading negative in $- \frac{27}{8}$ into the numerator.

$h'' (- \frac{3}{2}) = 20 (\frac{- 27}{8}) + 60 {(- \frac{3}{2})}^{2}$

Step 6.2.1.5.2

Factor $4$ out of $20$ .

$h'' (- \frac{3}{2}) = 4 (5) (\frac{- 27}{8}) + 60 {(- \frac{3}{2})}^{2}$

Step 6.2.1.5.3

Factor $4$ out of $8$ .

$h'' (- \frac{3}{2}) = 4 \cdot (5 (\frac{- 27}{4 \cdot 2})) + 60 {(- \frac{3}{2})}^{2}$

Step 6.2.1.5.4

Cancel the common factor.

$h'' (- \frac{3}{2}) = 4 \cdot (5 (\frac{- 27}{4 \cdot 2})) + 60 {(- \frac{3}{2})}^{2}$

Step 6.2.1.5.5

Rewrite the expression.

$h'' (- \frac{3}{2}) = 5 (\frac{- 27}{2}) + 60 {(- \frac{3}{2})}^{2}$

Step 6.2.1.6

Combine $5$ and $\frac{- 27}{2}$ .

$h'' (- \frac{3}{2}) = \frac{5 \cdot - 27}{2} + 60 {(- \frac{3}{2})}^{2}$

Step 6.2.1.7

Multiply $5$ by $- 27$ .

$h'' (- \frac{3}{2}) = \frac{- 135}{2} + 60 {(- \frac{3}{2})}^{2}$

Step 6.2.1.8

Move the negative in front of the fraction.

$h'' (- \frac{3}{2}) = - \frac{135}{2} + 60 {(- \frac{3}{2})}^{2}$

Step 6.2.1.9

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 6.2.1.9.1

Apply the product rule to $- \frac{3}{2}$ .

$h'' (- \frac{3}{2}) = - \frac{135}{2} + 60 ({(- 1)}^{2} {(\frac{3}{2})}^{2})$

Step 6.2.1.9.2

Apply the product rule to $\frac{3}{2}$ .

$h'' (- \frac{3}{2}) = - \frac{135}{2} + 60 ({(- 1)}^{2} (\frac{3^{2}}{2^{2}}))$

Step 6.2.1.10

Raise $- 1$ to the power of $2$ .

$h'' (- \frac{3}{2}) = - \frac{135}{2} + 60 (1 (\frac{3^{2}}{2^{2}}))$

Step 6.2.1.11

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

$h'' (- \frac{3}{2}) = - \frac{135}{2} + 60 (\frac{3^{2}}{2^{2}})$

Step 6.2.1.12

Raise $3$ to the power of $2$ .

$h'' (- \frac{3}{2}) = - \frac{135}{2} + 60 (\frac{9}{2^{2}})$

Step 6.2.1.13

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

$h'' (- \frac{3}{2}) = - \frac{135}{2} + 60 (\frac{9}{4})$

Step 6.2.1.14

Cancel the common factor of $4$ .

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Step 6.2.1.14.1

Factor $4$ out of $60$ .

$h'' (- \frac{3}{2}) = - \frac{135}{2} + 4 (15) (\frac{9}{4})$

Step 6.2.1.14.2

Cancel the common factor.

$h'' (- \frac{3}{2}) = - \frac{135}{2} + 4 \cdot (15 (\frac{9}{4}))$

Step 6.2.1.14.3

Rewrite the expression.

$h'' (- \frac{3}{2}) = - \frac{135}{2} + 15 \cdot 9$

Step 6.2.1.15

Multiply $15$ by $9$ .

$h'' (- \frac{3}{2}) = - \frac{135}{2} + 135$

Step 6.2.2

To write $135$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$h'' (- \frac{3}{2}) = - \frac{135}{2} + 135 \cdot \frac{2}{2}$

Step 6.2.3

Combine $135$ and $\frac{2}{2}$ .

$h'' (- \frac{3}{2}) = - \frac{135}{2} + \frac{135 \cdot 2}{2}$

Step 6.2.4

Combine the numerators over the common denominator.

$h'' (- \frac{3}{2}) = \frac{- 135 + 135 \cdot 2}{2}$

Step 6.2.5

Simplify the numerator.

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Step 6.2.5.1

Multiply $135$ by $2$ .

$h'' (- \frac{3}{2}) = \frac{- 135 + 270}{2}$

Step 6.2.5.2

Add $- 135$ and $270$ .

$h'' (- \frac{3}{2}) = \frac{135}{2}$

Step 6.2.6

The final answer is $\frac{135}{2}$ .

$\frac{135}{2}$

Step 6.3

At $- \frac{3}{2}$ , the second derivative is $67.5$ . Since this is positive, the second derivative is increasing on the interval $(- 3, 0)$ .

Increasing on $(- 3, 0)$ since $h'' (x) > 0$

Step 7

Substitute a value from the interval $(0, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $0.1$ in the expression.

$h'' (0.1) = 20 {(0.1)}^{3} + 60 {(0.1)}^{2}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Raise $0.1$ to the power of $3$ .

$h'' (0.1) = 20 \cdot 0.001 + 60 {(0.1)}^{2}$

Step 7.2.1.2

Multiply $20$ by $0.001$ .

$h'' (0.1) = 0.02 + 60 {(0.1)}^{2}$

Step 7.2.1.3

Raise $0.1$ to the power of $2$ .

$h'' (0.1) = 0.02 + 60 \cdot 0.01$

Step 7.2.1.4

Multiply $60$ by $0.01$ .

$h'' (0.1) = 0.02 + 0.6$

Step 7.2.2

Add $0.02$ and $0.6$ .

$h'' (0.1) = 0.62$

Step 7.2.3

The final answer is $0.62$ .

$0.62$

Step 7.3

At $0.1$ , the second derivative is $0.62$ . Since this is positive, the second derivative is increasing on the interval $(0, \infty)$ .

Increasing on $(0, \infty)$ since $h'' (x) > 0$

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(- 3, 162)$ .

$(- 3, 162)$

Step 9