Calculus Examples

$4 x^{5} \ln (6 x^{2})$

Step 1

Find the asymptotes.

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Step 1.1

Set the argument of the logarithm equal to zero.

$1296 x^{8} = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Divide each term in $1296 x^{8} = 0$ by $1296$ and simplify.

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Step 1.2.1.1

Divide each term in $1296 x^{8} = 0$ by $1296$ .

$\frac{1296 x^{8}}{1296} = \frac{0}{1296}$

Step 1.2.1.2

Simplify the left side.

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Step 1.2.1.2.1

Cancel the common factor of $1296$ .

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Step 1.2.1.2.1.1

Cancel the common factor.

$\frac{1296 x^{8}}{1296} = \frac{0}{1296}$

Step 1.2.1.2.1.2

Divide $x^{8}$ by $1$ .

$x^{8} = \frac{0}{1296}$

Step 1.2.1.3

Simplify the right side.

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Step 1.2.1.3.1

Divide $0$ by $1296$ .

$x^{8} = 0$

Step 1.2.2

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

$x = \pm \sqrt[8]{0}$

Step 1.2.3

Simplify $\pm \sqrt[8]{0}$ .

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Step 1.2.3.1

Rewrite $0$ as $0^{8}$ .

$x = \pm \sqrt[8]{0^{8}}$

Step 1.2.3.2

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

$x = \pm 0$

Step 1.2.3.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 1.3

The vertical asymptote occurs at $x = 0$ .

Vertical Asymptote: $x = 0$

Step 2

Find the point at $x = 1$ .

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Step 2.1

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

$f (1) = 4 {(1)}^{5} \ln (6 {(1)}^{2})$

Step 2.2

Simplify the result.

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Step 2.2.1

One to any power is one.

$f (1) = 4 \cdot (1 \ln (6 {(1)}^{2}))$

Step 2.2.2

Multiply $4$ by $1$ .

$f (1) = 4 \ln (6 {(1)}^{2})$

Step 2.2.3

One to any power is one.

$f (1) = 4 \ln (6 \cdot 1)$

Step 2.2.4

Multiply $6$ by $1$ .

$f (1) = 4 \ln (6)$

Step 2.2.5

Simplify $4 \ln (6)$ by moving $4$ inside the logarithm.

$f (1) = \ln (6^{4})$

Step 2.2.6

Raise $6$ to the power of $4$ .

$f (1) = \ln (1296)$

Step 2.2.7

The final answer is $\ln (1296)$ .

$\ln (1296)$

Step 2.3

Convert $\ln (1296)$ to decimal.

$y = 7.16703787$

Step 3

Find the point at $x = 2$ .

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Step 3.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = 4 {(2)}^{5} \ln (6 {(2)}^{2})$

Step 3.2

Simplify the result.

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Step 3.2.1

Raise $2$ to the power of $5$ .

$f (2) = 4 \cdot (32 \ln (6 {(2)}^{2}))$

Step 3.2.2

Multiply $4$ by $32$ .

$f (2) = 128 \ln (6 {(2)}^{2})$

Step 3.2.3

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

$f (2) = 128 \ln (6 \cdot 4)$

Step 3.2.4

Multiply $6$ by $4$ .

$f (2) = 128 \ln (24)$

Step 3.2.5

Simplify $128 \ln (24)$ by moving $128$ inside the logarithm.

$f (2) = \ln (24^{128})$

Step 3.2.6

The final answer is $\ln (24^{128})$ .

$\ln (24^{128})$

Step 3.3

Convert $\ln (24^{128})$ to decimal.

$y = 406.79089028$

Step 4

Find the point at $x = 3$ .

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Step 4.1

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

$f (3) = 4 {(3)}^{5} \ln (6 {(3)}^{2})$

Step 4.2

Simplify the result.

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Step 4.2.1

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

$f (3) = 4 \cdot (243 \ln (6 {(3)}^{2}))$

Step 4.2.2

Multiply $4$ by $243$ .

$f (3) = 972 \ln (6 {(3)}^{2})$

Step 4.2.3

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

$f (3) = 972 \ln (6 \cdot 9)$

Step 4.2.4

Multiply $6$ by $9$ .

$f (3) = 972 \ln (54)$

Step 4.2.5

The final answer is $972 \ln (54)$ .

$972 \ln (54)$

Step 4.3

Convert $972 \ln (54)$ to decimal.

$y = 3877.29249326$

Step 5

The log function can be graphed using the vertical asymptote at $x = 0$ and the points $(1, 7.16703787), (2, 406.79089028), (3, 3877.29249326)$ .

Vertical Asymptote: $x = 0$

$\begin{array}{c} x & y \\ 1 & 7.167 \\ 2 & 406.791 \\ 3 & 3877.292 \end{array}$

Step 6