Calculus Examples

$f (x) = x^{3} (6 x + 8)$

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 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 $f (x) = x^{3}$ and $g (x) = 6 x + 8$ .

$x^{3} \frac{d}{d x} [6 x + 8] + (6 x + 8) \frac{d}{d x} [x^{3}]$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $6 x + 8$ with respect to $x$ is $\frac{d}{d x} [6 x] + \frac{d}{d x} [8]$ .

$x^{3} (\frac{d}{d x} [6 x] + \frac{d}{d x} [8]) + (6 x + 8) \frac{d}{d x} [x^{3}]$

Step 1.1.2.2

Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$x^{3} (6 \frac{d}{d x} [x] + \frac{d}{d x} [8]) + (6 x + 8) \frac{d}{d x} [x^{3}]$

Step 1.1.2.3

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

$x^{3} (6 \cdot 1 + \frac{d}{d x} [8]) + (6 x + 8) \frac{d}{d x} [x^{3}]$

Step 1.1.2.4

Multiply $6$ by $1$ .

$x^{3} (6 + \frac{d}{d x} [8]) + (6 x + 8) \frac{d}{d x} [x^{3}]$

Step 1.1.2.5

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$x^{3} (6 + 0) + (6 x + 8) \frac{d}{d x} [x^{3}]$

Step 1.1.2.6

Simplify the expression.

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Step 1.1.2.6.1

Add $6$ and $0$ .

$x^{3} \cdot 6 + (6 x + 8) \frac{d}{d x} [x^{3}]$

Step 1.1.2.6.2

Move $6$ to the left of $x^{3}$ .

$6 \cdot x^{3} + (6 x + 8) \frac{d}{d x} [x^{3}]$

Step 1.1.2.7

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

$6 x^{3} + (6 x + 8) (3 x^{2})$

Step 1.1.2.8

Move $3$ to the left of $6 x + 8$ .

$6 x^{3} + 3 \cdot (6 x + 8) x^{2}$

Step 1.1.3

Simplify.

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Step 1.1.3.1

Apply the distributive property.

$6 x^{3} + (3 (6 x) + 3 \cdot 8) x^{2}$

Step 1.1.3.2

Apply the distributive property.

$6 x^{3} + 3 (6 x) x^{2} + 3 \cdot 8 x^{2}$

Step 1.1.3.3

Combine terms.

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Step 1.1.3.3.1

Multiply $6$ by $3$ .

$6 x^{3} + 18 x \cdot x^{2} + 3 \cdot 8 x^{2}$

Step 1.1.3.3.2

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 1.1.3.3.2.1

Move $x^{2}$ .

$6 x^{3} + 18 (x^{2} x) + 3 \cdot 8 x^{2}$

Step 1.1.3.3.2.2

Multiply $x^{2}$ by $x$ .

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Step 1.1.3.3.2.2.1

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

$6 x^{3} + 18 (x^{2} x^{1}) + 3 \cdot 8 x^{2}$

Step 1.1.3.3.2.2.2

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

$6 x^{3} + 18 x^{2 + 1} + 3 \cdot 8 x^{2}$

Step 1.1.3.3.2.3

Add $2$ and $1$ .

$6 x^{3} + 18 x^{3} + 3 \cdot 8 x^{2}$

Step 1.1.3.3.3

Multiply $3$ by $8$ .

$6 x^{3} + 18 x^{3} + 24 x^{2}$

Step 1.1.3.3.4

Add $6 x^{3}$ and $18 x^{3}$ .

$f' (x) = 24 x^{3} + 24 x^{2}$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $24 x^{3} + 24 x^{2}$ with respect to $x$ is $\frac{d}{d x} [24 x^{3}] + \frac{d}{d x} [24 x^{2}]$ .

$\frac{d}{d x} [24 x^{3}] + \frac{d}{d x} [24 x^{2}]$

Step 1.2.2

Evaluate $\frac{d}{d x} [24 x^{3}]$ .

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Step 1.2.2.1

Since $24$ is constant with respect to $x$ , the derivative of $24 x^{3}$ with respect to $x$ is $24 \frac{d}{d x} [x^{3}]$ .

$24 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [24 x^{2}]$

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 $n = 3$ .

$24 (3 x^{2}) + \frac{d}{d x} [24 x^{2}]$

Step 1.2.2.3

Multiply $3$ by $24$ .

$72 x^{2} + \frac{d}{d x} [24 x^{2}]$

Step 1.2.3

Evaluate $\frac{d}{d x} [24 x^{2}]$ .

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Step 1.2.3.1

Since $24$ is constant with respect to $x$ , the derivative of $24 x^{2}$ with respect to $x$ is $24 \frac{d}{d x} [x^{2}]$ .

$72 x^{2} + 24 \frac{d}{d x} [x^{2}]$

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 $n = 2$ .

$72 x^{2} + 24 (2 x)$

Step 1.2.3.3

Multiply $2$ by $24$ .

$f'' (x) = 72 x^{2} + 48 x$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $72 x^{2} + 48 x$ .

$72 x^{2} + 48 x$

Step 2

Set the second derivative equal to $0$ then solve the equation $72 x^{2} + 48 x = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$72 x^{2} + 48 x = 0$

Step 2.2

Factor $24 x$ out of $72 x^{2} + 48 x$ .

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Step 2.2.1

Factor $24 x$ out of $72 x^{2}$ .

$24 x (3 x) + 48 x = 0$

Step 2.2.2

Factor $24 x$ out of $48 x$ .

$24 x (3 x) + 24 x (2) = 0$

Step 2.2.3

Factor $24 x$ out of $24 x (3 x) + 24 x (2)$ .

$24 x (3 x + 2) = 0$

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 $0$ .

$x = 0$

$3 x + 2 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

Set $3 x + 2$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $3 x + 2$ equal to $0$ .

$3 x + 2 = 0$

Step 2.5.2

Solve $3 x + 2 = 0$ for $x$ .

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Step 2.5.2.1

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Step 2.5.2.2

Divide each term in $3 x = - 2$ by $3$ and simplify.

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Step 2.5.2.2.1

Divide each term in $3 x = - 2$ by $3$ .

$\frac{3 x}{3} = \frac{- 2}{3}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $3$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 2}{3}$

Step 2.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 2}{3}$

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{2}{3}$

Step 2.6

The final solution is all the values that make $24 x (3 x + 2) = 0$ true.

$x = 0, - \frac{2}{3}$

Step 3

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

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Step 3.1

Substitute $0$ in $f (x) = x^{3} (6 x + 8)$ to find the value of $y$ .

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Step 3.1.1

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

$f (0) = {(0)}^{3} (6 (0) + 8)$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

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

$f (0) = 0 (6 (0) + 8)$

Step 3.1.2.2

Multiply $6$ by $0$ .

$f (0) = 0 (0 + 8)$

Step 3.1.2.3

Add $0$ and $8$ .

$f (0) = 0 \cdot 8$

Step 3.1.2.4

Multiply $0$ by $8$ .

$f (0) = 0$

Step 3.1.2.5

The final answer is $0$ .

$0$

Step 3.2

The point found by substituting $0$ in $f (x) = x^{3} (6 x + 8)$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Step 3.3

Substitute $- \frac{2}{3}$ in $f (x) = x^{3} (6 x + 8)$ to find the value of $y$ .

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Step 3.3.1

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

$f (- \frac{2}{3}) = {(- \frac{2}{3})}^{3} (6 (- \frac{2}{3}) + 8)$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

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

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Step 3.3.2.1.1

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

$f (- \frac{2}{3}) = {(- 1)}^{3} {(\frac{2}{3})}^{3} (6 (- \frac{2}{3}) + 8)$

Step 3.3.2.1.2

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

$f (- \frac{2}{3}) = {(- 1)}^{3} (\frac{2^{3}}{3^{3}}) \cdot (6 (- \frac{2}{3}) + 8)$

Step 3.3.2.2

Evaluate the exponents.

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Step 3.3.2.2.1

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

$f (- \frac{2}{3}) = - \frac{2^{3}}{3^{3}} \cdot (6 (- \frac{2}{3}) + 8)$

Step 3.3.2.2.2

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

$f (- \frac{2}{3}) = - \frac{8}{3^{3}} \cdot (6 (- \frac{2}{3}) + 8)$

Step 3.3.2.2.3

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

$f (- \frac{2}{3}) = - \frac{8}{27} \cdot (6 (- \frac{2}{3}) + 8)$

Step 3.3.2.3

Simplify each term.

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Step 3.3.2.3.1

Cancel the common factor of $3$ .

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Step 3.3.2.3.1.1

Move the leading negative in $- \frac{2}{3}$ into the numerator.

$f (- \frac{2}{3}) = - \frac{8}{27} \cdot (6 (\frac{- 2}{3}) + 8)$

Step 3.3.2.3.1.2

Factor $3$ out of $6$ .

$f (- \frac{2}{3}) = - \frac{8}{27} \cdot (3 (2) (\frac{- 2}{3}) + 8)$

Step 3.3.2.3.1.3

Cancel the common factor.

$f (- \frac{2}{3}) = - \frac{8}{27} \cdot (3 \cdot (2 (\frac{- 2}{3})) + 8)$

Step 3.3.2.3.1.4

Rewrite the expression.

$f (- \frac{2}{3}) = - \frac{8}{27} \cdot (2 \cdot - 2 + 8)$

Step 3.3.2.3.2

Multiply $2$ by $- 2$ .

$f (- \frac{2}{3}) = - \frac{8}{27} \cdot (- 4 + 8)$

Step 3.3.2.4

Add $- 4$ and $8$ .

$f (- \frac{2}{3}) = - \frac{8}{27} \cdot 4$

Step 3.3.2.5

Multiply $- \frac{8}{27} \cdot 4$ .

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Step 3.3.2.5.1

Multiply $4$ by $- 1$ .

$f (- \frac{2}{3}) = - 4 (\frac{8}{27})$

Step 3.3.2.5.2

Combine $- 4$ and $\frac{8}{27}$ .

$f (- \frac{2}{3}) = \frac{- 4 \cdot 8}{27}$

Step 3.3.2.5.3

Multiply $- 4$ by $8$ .

$f (- \frac{2}{3}) = \frac{- 32}{27}$

Step 3.3.2.6

Move the negative in front of the fraction.

$f (- \frac{2}{3}) = - \frac{32}{27}$

Step 3.3.2.7

The final answer is $- \frac{32}{27}$ .

$- \frac{32}{27}$

Step 3.4

The point found by substituting $- \frac{2}{3}$ in $f (x) = x^{3} (6 x + 8)$ is $(- \frac{2}{3}, - \frac{32}{27})$ . This point can be an inflection point.

$(- \frac{2}{3}, - \frac{32}{27})$

Step 3.5

Determine the points that could be inflection points.

$(0, 0), (- \frac{2}{3}, - \frac{32}{27})$

Step 4

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

$(- \infty, - \frac{2}{3}) \cup (- \frac{2}{3}, 0) \cup (0, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - \frac{2}{3})$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 0.7\overline{6}$ in the expression.

$f'' (- 0.7\overline{6}) = 72 {(- 0.7\overline{6})}^{2} + 48 (- 0.7\overline{6})$

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 $- 0.7\overline{6}$ to the power of $2$ .

$f'' (- 0.7\overline{6}) = 72 \cdot 0.58\overline{7} + 48 (- 0.7\overline{6})$

Step 5.2.1.2

Multiply $72$ by $0.58\overline{7}$ .

$f'' (- 0.7\overline{6}) = 42.32 + 48 (- 0.7\overline{6})$

Step 5.2.1.3

Multiply $48$ by $- 0.7\overline{6}$ .

$f'' (- 0.7\overline{6}) = 42.32 - 36.8$

Step 5.2.2

Subtract $36.8$ from $42.32$ .

$f'' (- 0.7\overline{6}) = 5.52$

Step 5.2.3

The final answer is $5.52$ .

$5.52$

Step 5.3

At $- 0.7\overline{6}$ , the second derivative is $5.52$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, - \frac{2}{3})$ .

Increasing on $(- \infty, - \frac{2}{3})$ since $f'' (x) > 0$

Step 6

Substitute a value from the interval $(- \frac{2}{3}, 0)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 0.\overline{3}$ in the expression.

$f'' (- 0.\overline{3}) = 72 {(- 0.\overline{3})}^{2} + 48 (- 0.\overline{3})$

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 $- 0.\overline{3}$ to the power of $2$ .

$f'' (- 0.\overline{3}) = 72 \cdot 0.\overline{1} + 48 (- 0.\overline{3})$

Step 6.2.1.2

Multiply $72$ by $0.\overline{1}$ .

$f'' (- 0.\overline{3}) = 8 + 48 (- 0.\overline{3})$

Step 6.2.1.3

Multiply $48$ by $- 0.\overline{3}$ .

$f'' (- 0.\overline{3}) = 8 - 16$

Step 6.2.2

Subtract $16$ from $8$ .

$f'' (- 0.\overline{3}) = - 8$

Step 6.2.3

The final answer is $- 8$ .

$- 8$

Step 6.3

At $- 0.\overline{3}$ , the second derivative is $- 8$ . Since this is negative, the second derivative is decreasing on the interval $(- \frac{2}{3}, 0)$

Decreasing on $(- \frac{2}{3}, 0)$ since $f'' (x) < 0$

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.

$f'' (0.1) = 72 {(0.1)}^{2} + 48 (0.1)$

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 $2$ .

$f'' (0.1) = 72 \cdot 0.01 + 48 (0.1)$

Step 7.2.1.2

Multiply $72$ by $0.01$ .

$f'' (0.1) = 0.72 + 48 (0.1)$

Step 7.2.1.3

Multiply $48$ by $0.1$ .

$f'' (0.1) = 0.72 + 4.8$

Step 7.2.2

Add $0.72$ and $4.8$ .

$f'' (0.1) = 5.52$

Step 7.2.3

The final answer is $5.52$ .

$5.52$

Step 7.3

At $0.1$ , the second derivative is $5.52$ . Since this is positive, the second derivative is increasing on the interval $(0, \infty)$ .

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

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 points in this case are $(- \frac{2}{3}, - \frac{32}{27}), (0, 0)$ .

$(- \frac{2}{3}, - \frac{32}{27}), (0, 0)$

Step 9