Calculus Examples

$y = 5 x^{4} + 3 x^{5}$

Step 1

Write $y = 5 x^{4} + 3 x^{5}$ as a function.

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

Step 2

Find the second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

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

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

Step 2.1.2

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

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Step 2.1.2.1

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

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

Step 2.1.2.2

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

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

Step 2.1.2.3

Multiply $4$ by $5$ .

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

Step 2.1.3

Evaluate $\frac{d}{d x} [3 x^{5}]$ .

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Step 2.1.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{5}$ with respect to $x$ is $3 \frac{d}{d x} [x^{5}]$ .

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

Step 2.1.3.2

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

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

Step 2.1.3.3

Multiply $5$ by $3$ .

$20 x^{3} + 15 x^{4}$

Step 2.1.4

Reorder terms.

$f' (x) = 15 x^{4} + 20 x^{3}$

Step 2.2

Find the second derivative.

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Step 2.2.1

By the Sum Rule, the derivative of $15 x^{4} + 20 x^{3}$ with respect to $x$ is $\frac{d}{d x} [15 x^{4}] + \frac{d}{d x} [20 x^{3}]$ .

$\frac{d}{d x} [15 x^{4}] + \frac{d}{d x} [20 x^{3}]$

Step 2.2.2

Evaluate $\frac{d}{d x} [15 x^{4}]$ .

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Step 2.2.2.1

Since $15$ is constant with respect to $x$ , the derivative of $15 x^{4}$ with respect to $x$ is $15 \frac{d}{d x} [x^{4}]$ .

$15 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [20 x^{3}]$

Step 2.2.2.2

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

$15 (4 x^{3}) + \frac{d}{d x} [20 x^{3}]$

Step 2.2.2.3

Multiply $4$ by $15$ .

$60 x^{3} + \frac{d}{d x} [20 x^{3}]$

Step 2.2.3

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

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Step 2.2.3.1

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

$60 x^{3} + 20 \frac{d}{d x} [x^{3}]$

Step 2.2.3.2

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

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

Step 2.2.3.3

Multiply $3$ by $20$ .

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

Step 2.3

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

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

Step 3

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

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Step 3.1

Set the second derivative equal to $0$ .

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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

$60 x^{2} (x) + 60 x^{2} (1) = 0$

Step 3.2.3

Factor $60 x^{2}$ out of $60 x^{2} (x) + 60 x^{2} (1)$ .

$60 x^{2} (x + 1) = 0$

Step 3.3

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

$x^{2} = 0$

$x + 1 = 0$

Step 3.4

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

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Step 3.4.1

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

$x^{2} = 0$

Step 3.4.2

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

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Step 3.4.2.1

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

$x = \pm \sqrt{0}$

Step 3.4.2.2

Simplify $\pm \sqrt{0}$ .

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Step 3.4.2.2.1

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

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

Step 3.4.2.2.2

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

$x = \pm 0$

Step 3.4.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 3.5

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 3.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.6

The final solution is all the values that make $60 x^{2} (x + 1) = 0$ true.

$x = 0, - 1$

Step 4

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

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Step 4.1

Substitute $0$ in $f (x) = 5 x^{4} + 3 x^{5}$ to find the value of $y$ .

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Step 4.1.1

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

$f (0) = 5 {(0)}^{4} + 3 {(0)}^{5}$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Simplify each term.

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Step 4.1.2.1.1

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

$f (0) = 5 \cdot 0 + 3 {(0)}^{5}$

Step 4.1.2.1.2

Multiply $5$ by $0$ .

$f (0) = 0 + 3 {(0)}^{5}$

Step 4.1.2.1.3

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

$f (0) = 0 + 3 \cdot 0$

Step 4.1.2.1.4

Multiply $3$ by $0$ .

$f (0) = 0 + 0$

Step 4.1.2.2

Add $0$ and $0$ .

$f (0) = 0$

Step 4.1.2.3

The final answer is $0$ .

$0$

Step 4.2

The point found by substituting $0$ in $f (x) = 5 x^{4} + 3 x^{5}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Step 4.3

Substitute $- 1$ in $f (x) = 5 x^{4} + 3 x^{5}$ to find the value of $y$ .

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Step 4.3.1

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = 5 {(- 1)}^{4} + 3 {(- 1)}^{5}$

Step 4.3.2

Simplify the result.

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Step 4.3.2.1

Simplify each term.

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Step 4.3.2.1.1

Raise $- 1$ to the power of $4$ .

$f (- 1) = 5 \cdot 1 + 3 {(- 1)}^{5}$

Step 4.3.2.1.2

Multiply $5$ by $1$ .

$f (- 1) = 5 + 3 {(- 1)}^{5}$

Step 4.3.2.1.3

Raise $- 1$ to the power of $5$ .

$f (- 1) = 5 + 3 \cdot - 1$

Step 4.3.2.1.4

Multiply $3$ by $- 1$ .

$f (- 1) = 5 - 3$

Step 4.3.2.2

Subtract $3$ from $5$ .

$f (- 1) = 2$

Step 4.3.2.3

The final answer is $2$ .

$2$

Step 4.4

The point found by substituting $- 1$ in $f (x) = 5 x^{4} + 3 x^{5}$ is $(- 1, 2)$ . This point can be an inflection point.

$(- 1, 2)$

Step 4.5

Determine the points that could be inflection points.

$(0, 0), (- 1, 2)$

Step 5

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

$(- \infty, - 1) \cup (- 1, 0) \cup (0, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - 1)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 1.1$ in the expression.

$f'' (- 1.1) = 60 {(- 1.1)}^{3} + 60 {(- 1.1)}^{2}$

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

Raise $- 1.1$ to the power of $3$ .

$f'' (- 1.1) = 60 \cdot - 1.331 + 60 {(- 1.1)}^{2}$

Step 6.2.1.2

Multiply $60$ by $- 1.331$ .

$f'' (- 1.1) = - 79.86 + 60 {(- 1.1)}^{2}$

Step 6.2.1.3

Raise $- 1.1$ to the power of $2$ .

$f'' (- 1.1) = - 79.86 + 60 \cdot 1.21$

Step 6.2.1.4

Multiply $60$ by $1.21$ .

$f'' (- 1.1) = - 79.86 + 72.6$

Step 6.2.2

Add $- 79.86$ and $72.6$ .

$f'' (- 1.1) = - 7.26$

Step 6.2.3

The final answer is $- 7.26$ .

$- 7.26$

Step 6.3

At $- 1.1$ , the second derivative is $- 7.26$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, - 1)$

Decreasing on $(- \infty, - 1)$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(- 1, 0)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $- \frac{1}{2}$ in the expression.

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

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

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

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Step 7.2.1.1.1

Apply the product rule to $- \frac{1}{2}$ .

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

Step 7.2.1.1.2

Apply the product rule to $\frac{1}{2}$ .

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

Step 7.2.1.2

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

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

Step 7.2.1.3

One to any power is one.

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

Step 7.2.1.4

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

$f'' (- \frac{1}{2}) = 60 (- \frac{1}{8}) + 60 {(- \frac{1}{2})}^{2}$

Step 7.2.1.5

Cancel the common factor of $4$ .

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Step 7.2.1.5.1

Move the leading negative in $- \frac{1}{8}$ into the numerator.

$f'' (- \frac{1}{2}) = 60 (\frac{- 1}{8}) + 60 {(- \frac{1}{2})}^{2}$

Step 7.2.1.5.2

Factor $4$ out of $60$ .

$f'' (- \frac{1}{2}) = 4 (15) (\frac{- 1}{8}) + 60 {(- \frac{1}{2})}^{2}$

Step 7.2.1.5.3

Factor $4$ out of $8$ .

$f'' (- \frac{1}{2}) = 4 \cdot (15 (\frac{- 1}{4 \cdot 2})) + 60 {(- \frac{1}{2})}^{2}$

Step 7.2.1.5.4

Cancel the common factor.

$f'' (- \frac{1}{2}) = 4 \cdot (15 (\frac{- 1}{4 \cdot 2})) + 60 {(- \frac{1}{2})}^{2}$

Step 7.2.1.5.5

Rewrite the expression.

$f'' (- \frac{1}{2}) = 15 (\frac{- 1}{2}) + 60 {(- \frac{1}{2})}^{2}$

Step 7.2.1.6

Combine $15$ and $\frac{- 1}{2}$ .

$f'' (- \frac{1}{2}) = \frac{15 \cdot - 1}{2} + 60 {(- \frac{1}{2})}^{2}$

Step 7.2.1.7

Multiply $15$ by $- 1$ .

$f'' (- \frac{1}{2}) = \frac{- 15}{2} + 60 {(- \frac{1}{2})}^{2}$

Step 7.2.1.8

Move the negative in front of the fraction.

$f'' (- \frac{1}{2}) = - \frac{15}{2} + 60 {(- \frac{1}{2})}^{2}$

Step 7.2.1.9

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

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Step 7.2.1.9.1

Apply the product rule to $- \frac{1}{2}$ .

$f'' (- \frac{1}{2}) = - \frac{15}{2} + 60 ({(- 1)}^{2} {(\frac{1}{2})}^{2})$

Step 7.2.1.9.2

Apply the product rule to $\frac{1}{2}$ .

$f'' (- \frac{1}{2}) = - \frac{15}{2} + 60 ({(- 1)}^{2} (\frac{1^{2}}{2^{2}}))$

Step 7.2.1.10

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

$f'' (- \frac{1}{2}) = - \frac{15}{2} + 60 (1 (\frac{1^{2}}{2^{2}}))$

Step 7.2.1.11

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

$f'' (- \frac{1}{2}) = - \frac{15}{2} + 60 (\frac{1^{2}}{2^{2}})$

Step 7.2.1.12

One to any power is one.

$f'' (- \frac{1}{2}) = - \frac{15}{2} + 60 (\frac{1}{2^{2}})$

Step 7.2.1.13

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

$f'' (- \frac{1}{2}) = - \frac{15}{2} + 60 (\frac{1}{4})$

Step 7.2.1.14

Cancel the common factor of $4$ .

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Step 7.2.1.14.1

Factor $4$ out of $60$ .

$f'' (- \frac{1}{2}) = - \frac{15}{2} + 4 (15) (\frac{1}{4})$

Step 7.2.1.14.2

Cancel the common factor.

$f'' (- \frac{1}{2}) = - \frac{15}{2} + 4 \cdot (15 (\frac{1}{4}))$

Step 7.2.1.14.3

Rewrite the expression.

$f'' (- \frac{1}{2}) = - \frac{15}{2} + 15$

Step 7.2.2

To write $15$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f'' (- \frac{1}{2}) = - \frac{15}{2} + 15 \cdot \frac{2}{2}$

Step 7.2.3

Combine $15$ and $\frac{2}{2}$ .

$f'' (- \frac{1}{2}) = - \frac{15}{2} + \frac{15 \cdot 2}{2}$

Step 7.2.4

Combine the numerators over the common denominator.

$f'' (- \frac{1}{2}) = \frac{- 15 + 15 \cdot 2}{2}$

Step 7.2.5

Simplify the numerator.

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Step 7.2.5.1

Multiply $15$ by $2$ .

$f'' (- \frac{1}{2}) = \frac{- 15 + 30}{2}$

Step 7.2.5.2

Add $- 15$ and $30$ .

$f'' (- \frac{1}{2}) = \frac{15}{2}$

Step 7.2.6

The final answer is $\frac{15}{2}$ .

$\frac{15}{2}$

Step 7.3

At $- \frac{1}{2}$ , the second derivative is $7.5$ . Since this is positive, the second derivative is increasing on the interval $(- 1, 0)$ .

Increasing on $(- 1, 0)$ since $f'' (x) > 0$

Step 8

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 8.1

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

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

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

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

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

Step 8.2.1.2

Multiply $60$ by $0.001$ .

$f'' (0.1) = 0.06 + 60 {(0.1)}^{2}$

Step 8.2.1.3

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

$f'' (0.1) = 0.06 + 60 \cdot 0.01$

Step 8.2.1.4

Multiply $60$ by $0.01$ .

$f'' (0.1) = 0.06 + 0.6$

Step 8.2.2

Add $0.06$ and $0.6$ .

$f'' (0.1) = 0.66$

Step 8.2.3

The final answer is $0.66$ .

$0.66$

Step 8.3

At $0.1$ , the second derivative is $0.66$ . Since this is positive, the second derivative is increasing on the interval $(0, \infty)$ .

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

Step 9

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 $(- 1, 2)$ .

$(- 1, 2)$

Step 10