Calculus Examples

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

Step 1

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

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

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 $3 x^{4} + 4 x^{3}$ with respect to $x$ is $\frac{d}{d x} [3 x^{4}] + \frac{d}{d x} [4 x^{3}]$ .

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

Step 2.1.2

Evaluate $\frac{d}{d x} [3 x^{4}]$ .

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Step 2.1.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{4}$ with respect to $x$ is $3 \frac{d}{d x} [x^{4}]$ .

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

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 $n = 4$ .

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

Step 2.1.2.3

Multiply $4$ by $3$ .

$12 x^{3} + \frac{d}{d x} [4 x^{3}]$

Step 2.1.3

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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Step 2.1.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3}$ with respect to $x$ is $4 \frac{d}{d x} [x^{3}]$ .

$12 x^{3} + 4 \frac{d}{d x} [x^{3}]$

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 $n = 3$ .

$12 x^{3} + 4 (3 x^{2})$

Step 2.1.3.3

Multiply $3$ by $4$ .

$f' (x) = 12 x^{3} + 12 x^{2}$

Step 2.2

Find the second derivative.

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Step 2.2.1

By the Sum Rule, the derivative of $12 x^{3} + 12 x^{2}$ with respect to $x$ is $\frac{d}{d x} [12 x^{3}] + \frac{d}{d x} [12 x^{2}]$ .

$\frac{d}{d x} [12 x^{3}] + \frac{d}{d x} [12 x^{2}]$

Step 2.2.2

Evaluate $\frac{d}{d x} [12 x^{3}]$ .

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Step 2.2.2.1

Since $12$ is constant with respect to $x$ , the derivative of $12 x^{3}$ with respect to $x$ is $12 \frac{d}{d x} [x^{3}]$ .

$12 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [12 x^{2}]$

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 $n = 3$ .

$12 (3 x^{2}) + \frac{d}{d x} [12 x^{2}]$

Step 2.2.2.3

Multiply $3$ by $12$ .

$36 x^{2} + \frac{d}{d x} [12 x^{2}]$

Step 2.2.3

Evaluate $\frac{d}{d x} [12 x^{2}]$ .

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Step 2.2.3.1

Since $12$ is constant with respect to $x$ , the derivative of $12 x^{2}$ with respect to $x$ is $12 \frac{d}{d x} [x^{2}]$ .

$36 x^{2} + 12 \frac{d}{d x} [x^{2}]$

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 $n = 2$ .

$36 x^{2} + 12 (2 x)$

Step 2.2.3.3

Multiply $2$ by $12$ .

$f'' (x) = 36 x^{2} + 24 x$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $36 x^{2} + 24 x$ .

$36 x^{2} + 24 x$

Step 3

Set the second derivative equal to $0$ then solve the equation $36 x^{2} + 24 x = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$36 x^{2} + 24 x = 0$

Step 3.2

Factor $12 x$ out of $36 x^{2} + 24 x$ .

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Step 3.2.1

Factor $12 x$ out of $36 x^{2}$ .

$12 x (3 x) + 24 x = 0$

Step 3.2.2

Factor $12 x$ out of $24 x$ .

$12 x (3 x) + 12 x (2) = 0$

Step 3.2.3

Factor $12 x$ out of $12 x (3 x) + 12 x (2)$ .

$12 x (3 x + 2) = 0$

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 $0$ .

$x = 0$

$3 x + 2 = 0$

Step 3.4

Set $x$ equal to $0$ .

$x = 0$

Step 3.5

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

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Step 3.5.1

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

$3 x + 2 = 0$

Step 3.5.2

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

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Step 3.5.2.1

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Step 3.5.2.2

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

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Step 3.5.2.2.1

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

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

Step 3.5.2.2.2

Simplify the left side.

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Step 3.5.2.2.2.1

Cancel the common factor of $3$ .

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Step 3.5.2.2.2.1.1

Cancel the common factor.

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

Step 3.5.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 3.5.2.2.3

Simplify the right side.

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Step 3.5.2.2.3.1

Move the negative in front of the fraction.

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

Step 3.6

The final solution is all the values that make $12 x (3 x + 2) = 0$ true.

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

Step 4

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

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Step 4.1

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

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Step 4.1.1

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

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

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 $0$ .

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

Step 4.1.2.1.2

Multiply $3$ by $0$ .

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

Step 4.1.2.1.3

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

$f (0) = 0 + 4 \cdot 0$

Step 4.1.2.1.4

Multiply $4$ 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 $f (x) = 3 x^{4} + 4 x^{3}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Step 4.3

Substitute $- \frac{2}{3}$ in $f (x) = 3 x^{4} + 4 x^{3}$ to find the value of $y$ .

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Step 4.3.1

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

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

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

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

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Step 4.3.2.1.1.1

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

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

Step 4.3.2.1.1.2

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

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

Step 4.3.2.1.2

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

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

Step 4.3.2.1.3

Multiply $\frac{2^{4}}{3^{4}}$ by $1$ .

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

Step 4.3.2.1.4

Raise $2$ to the power of $4$ .

$f (- \frac{2}{3}) = 3 (\frac{16}{3^{4}}) + 4 {(- \frac{2}{3})}^{3}$

Step 4.3.2.1.5

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

$f (- \frac{2}{3}) = 3 (\frac{16}{81}) + 4 {(- \frac{2}{3})}^{3}$

Step 4.3.2.1.6

Cancel the common factor of $3$ .

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Step 4.3.2.1.6.1

Factor $3$ out of $81$ .

$f (- \frac{2}{3}) = 3 (\frac{16}{3 (27)}) + 4 {(- \frac{2}{3})}^{3}$

Step 4.3.2.1.6.2

Cancel the common factor.

$f (- \frac{2}{3}) = 3 (\frac{16}{3 \cdot 27}) + 4 {(- \frac{2}{3})}^{3}$

Step 4.3.2.1.6.3

Rewrite the expression.

$f (- \frac{2}{3}) = \frac{16}{27} + 4 {(- \frac{2}{3})}^{3}$

Step 4.3.2.1.7

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

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Step 4.3.2.1.7.1

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

$f (- \frac{2}{3}) = \frac{16}{27} + 4 ({(- 1)}^{3} {(\frac{2}{3})}^{3})$

Step 4.3.2.1.7.2

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

$f (- \frac{2}{3}) = \frac{16}{27} + 4 ({(- 1)}^{3} (\frac{2^{3}}{3^{3}}))$

Step 4.3.2.1.8

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

$f (- \frac{2}{3}) = \frac{16}{27} + 4 (- \frac{2^{3}}{3^{3}})$

Step 4.3.2.1.9

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

$f (- \frac{2}{3}) = \frac{16}{27} + 4 (- \frac{8}{3^{3}})$

Step 4.3.2.1.10

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

$f (- \frac{2}{3}) = \frac{16}{27} + 4 (- \frac{8}{27})$

Step 4.3.2.1.11

Multiply $4 (- \frac{8}{27})$ .

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Step 4.3.2.1.11.1

Multiply $- 1$ by $4$ .

$f (- \frac{2}{3}) = \frac{16}{27} - 4 (\frac{8}{27})$

Step 4.3.2.1.11.2

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

$f (- \frac{2}{3}) = \frac{16}{27} + \frac{- 4 \cdot 8}{27}$

Step 4.3.2.1.11.3

Multiply $- 4$ by $8$ .

$f (- \frac{2}{3}) = \frac{16}{27} + \frac{- 32}{27}$

Step 4.3.2.1.12

Move the negative in front of the fraction.

$f (- \frac{2}{3}) = \frac{16}{27} - \frac{32}{27}$

Step 4.3.2.2

Combine fractions.

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Step 4.3.2.2.1

Combine the numerators over the common denominator.

$f (- \frac{2}{3}) = \frac{16 - 32}{27}$

Step 4.3.2.2.2

Simplify the expression.

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Step 4.3.2.2.2.1

Subtract $32$ from $16$ .

$f (- \frac{2}{3}) = \frac{- 16}{27}$

Step 4.3.2.2.2.2

Move the negative in front of the fraction.

$f (- \frac{2}{3}) = - \frac{16}{27}$

Step 4.3.2.3

The final answer is $- \frac{16}{27}$ .

$- \frac{16}{27}$

Step 4.4

The point found by substituting $- \frac{2}{3}$ in $f (x) = 3 x^{4} + 4 x^{3}$ is $(- \frac{2}{3}, - \frac{16}{27})$ . This point can be an inflection point.

$(- \frac{2}{3}, - \frac{16}{27})$

Step 4.5

Determine the points that could be inflection points.

$(0, 0), (- \frac{2}{3}, - \frac{16}{27})$

Step 5

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 6

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 6.1

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

$f'' (- 0.7\overline{6}) = 36 {(- 0.7\overline{6})}^{2} + 24 (- 0.7\overline{6})$

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

$f'' (- 0.7\overline{6}) = 36 \cdot 0.58\overline{7} + 24 (- 0.7\overline{6})$

Step 6.2.1.2

Multiply $36$ by $0.58\overline{7}$ .

$f'' (- 0.7\overline{6}) = 21.16 + 24 (- 0.7\overline{6})$

Step 6.2.1.3

Multiply $24$ by $- 0.7\overline{6}$ .

$f'' (- 0.7\overline{6}) = 21.16 - 18.4$

Step 6.2.2

Subtract $18.4$ from $21.16$ .

$f'' (- 0.7\overline{6}) = 2.76$

Step 6.2.3

The final answer is $2.76$ .

$2.76$

Step 6.3

At $- 0.7\overline{6}$ , the second derivative is $2.76$ . 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 7

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 7.1

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

$f'' (- 0.\overline{3}) = 36 {(- 0.\overline{3})}^{2} + 24 (- 0.\overline{3})$

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

$f'' (- 0.\overline{3}) = 36 \cdot 0.\overline{1} + 24 (- 0.\overline{3})$

Step 7.2.1.2

Multiply $36$ by $0.\overline{1}$ .

$f'' (- 0.\overline{3}) = 4 + 24 (- 0.\overline{3})$

Step 7.2.1.3

Multiply $24$ by $- 0.\overline{3}$ .

$f'' (- 0.\overline{3}) = 4 - 8$

Step 7.2.2

Subtract $8$ from $4$ .

$f'' (- 0.\overline{3}) = - 4$

Step 7.2.3

The final answer is $- 4$ .

$- 4$

Step 7.3

At $- 0.\overline{3}$ , the second derivative is $- 4$ . 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 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.

$f'' (0.1) = 36 {(0.1)}^{2} + 24 (0.1)$

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 $2$ .

$f'' (0.1) = 36 \cdot 0.01 + 24 (0.1)$

Step 8.2.1.2

Multiply $36$ by $0.01$ .

$f'' (0.1) = 0.36 + 24 (0.1)$

Step 8.2.1.3

Multiply $24$ by $0.1$ .

$f'' (0.1) = 0.36 + 2.4$

Step 8.2.2

Add $0.36$ and $2.4$ .

$f'' (0.1) = 2.76$

Step 8.2.3

The final answer is $2.76$ .

$2.76$

Step 8.3

At $0.1$ , the second derivative is $2.76$ . 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 points in this case are $(- \frac{2}{3}, - \frac{16}{27}), (0, 0)$ .

$(- \frac{2}{3}, - \frac{16}{27}), (0, 0)$

Step 10