Calculus Examples

$\frac{1}{3} \cdot ({(4 + \frac{1}{x})}^{3} (\ln (x)))$

Step 1

Find the asymptotes.

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Step 1.1

Find where the expression $\frac{\ln (x) {(4 x + 1)}^{3}}{3 x^{3}}$ is undefined.

$x \leq 0$

Step 1.2

Since $\frac{\ln (x) {(4 x + 1)}^{3}}{3 x^{3}}$ $\to$ $\infty$ as $x$ $\to$ $0$ from the left and $\frac{\ln (x) {(4 x + 1)}^{3}}{3 x^{3}}$ $\to$ $- \infty$ as $x$ $\to$ $0$ from the right, then $x = 0$ is a vertical asymptote.

$x = 0$

Step 1.3

Ignoring the logarithm, consider the rational function $R (x) = \frac{a x^{n}}{b x^{m}}$ where $n$ is the degree of the numerator and $m$ is the degree of the denominator.

1. If $n < m$ , then the x-axis, $y = 0$ , is the horizontal asymptote.

2. If $n = m$ , then the horizontal asymptote is the line $y = \frac{a}{b}$ .

3. If $n > m$ , then there is no horizontal asymptote (there is an oblique asymptote).

Step 1.4

Find $n$ and $m$ .

$n = 3$

$m = 3$

Step 1.5

Since $n = m$ , the horizontal asymptote is the line $y = \frac{a}{b}$ where $a = 64$ and $b = 3$ .

$y = \frac{64}{3}$

Step 1.6

No oblique asymptotes are present for logarithmic and trigonometric functions.

No Oblique Asymptotes

Step 1.7

This is the set of all asymptotes.

Vertical Asymptotes: $x = 0$

Horizontal Asymptotes: $y = \frac{64}{3}$

Vertical Asymptotes: $x = 0$

Horizontal Asymptotes: $y = \frac{64}{3}$

Step 2

Find the point at $x = 1$ .

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Step 2.1

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

$f (1) = \frac{1}{3} \cdot ({(4 + \frac{1}{1})}^{3} (\ln (1)))$

Step 2.2

Simplify the result.

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Step 2.2.1

Divide $1$ by $1$ .

$f (1) = \frac{1}{3} \cdot ({(4 + 1)}^{3} \ln (1))$

Step 2.2.2

Add $4$ and $1$ .

$f (1) = \frac{1}{3} \cdot (5^{3} \ln (1))$

Step 2.2.3

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

$f (1) = \frac{1}{3} \cdot (125 \ln (1))$

Step 2.2.4

The natural logarithm of $1$ is $0$ .

$f (1) = \frac{1}{3} \cdot (125 \cdot 0)$

Step 2.2.5

Multiply $125$ by $0$ .

$f (1) = \frac{1}{3} \cdot 0$

Step 2.2.6

Multiply $\frac{1}{3}$ by $0$ .

$f (1) = 0$

Step 2.2.7

The final answer is $0$ .

$0$

Step 2.3

Convert $0$ to decimal.

$y = 0$

Step 3

Find the point at $x = 2$ .

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Step 3.1

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

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

Step 3.2

Simplify the result.

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Step 3.2.1

To write $4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.2.2

Combine $4$ and $\frac{2}{2}$ .

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

Step 3.2.3

Combine the numerators over the common denominator.

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

Step 3.2.4

Simplify the numerator.

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Step 3.2.4.1

Multiply $4$ by $2$ .

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

Step 3.2.4.2

Add $8$ and $1$ .

$f (2) = \frac{1}{3} \cdot ({(\frac{9}{2})}^{3} \ln (2))$

Step 3.2.5

Apply the product rule to $\frac{9}{2}$ .

$f (2) = \frac{1}{3} \cdot (\frac{9^{3}}{2^{3}} \cdot \ln (2))$

Step 3.2.6

Raise $9$ to the power of $3$ .

$f (2) = \frac{1}{3} \cdot (\frac{729}{2^{3}} \cdot \ln (2))$

Step 3.2.7

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

$f (2) = \frac{1}{3} \cdot (\frac{729}{8} \cdot \ln (2))$

Step 3.2.8

Simplify $\frac{729}{8} \ln (2)$ by moving $\frac{729}{8}$ inside the logarithm.

$f (2) = \frac{1}{3} \cdot \ln (2^{\frac{729}{8}})$

Step 3.2.9

Simplify $\frac{1}{3} \ln (2^{\frac{729}{8}})$ by moving $\frac{1}{3}$ inside the logarithm.

$f (2) = \ln ({(2^{\frac{729}{8}})}^{\frac{1}{3}})$

Step 3.2.10

Multiply the exponents in ${(2^{\frac{729}{8}})}^{\frac{1}{3}}$ .

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Step 3.2.10.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (2) = \ln (2^{\frac{729}{8} \cdot \frac{1}{3}})$

Step 3.2.10.2

Cancel the common factor of $3$ .

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Step 3.2.10.2.1

Factor $3$ out of $729$ .

$f (2) = \ln (2^{\frac{3 (243)}{8} \cdot \frac{1}{3}})$

Step 3.2.10.2.2

Cancel the common factor.

$f (2) = \ln (2^{\frac{3 \cdot 243}{8} \cdot \frac{1}{3}})$

Step 3.2.10.2.3

Rewrite the expression.

$f (2) = \ln (2^{\frac{243}{8}})$

Step 3.2.11

The final answer is $\ln (2^{\frac{243}{8}})$ .

$\ln (2^{\frac{243}{8}})$

Step 3.3

Convert $\ln (2^{\frac{243}{8}})$ to decimal.

$y = 21.0543456$

Step 4

Find the point at $x = 3$ .

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Step 4.1

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

$f (3) = \frac{1}{3} \cdot ({(4 + \frac{1}{3})}^{3} (\ln (3)))$

Step 4.2

Simplify the result.

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Step 4.2.1

To write $4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f (3) = \frac{1}{3} \cdot ({(4 \cdot \frac{3}{3} + \frac{1}{3})}^{3} \ln (3))$

Step 4.2.2

Combine $4$ and $\frac{3}{3}$ .

$f (3) = \frac{1}{3} \cdot ({(\frac{4 \cdot 3}{3} + \frac{1}{3})}^{3} \ln (3))$

Step 4.2.3

Combine the numerators over the common denominator.

$f (3) = \frac{1}{3} \cdot ({(\frac{4 \cdot 3 + 1}{3})}^{3} \ln (3))$

Step 4.2.4

Simplify the numerator.

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Step 4.2.4.1

Multiply $4$ by $3$ .

$f (3) = \frac{1}{3} \cdot ({(\frac{12 + 1}{3})}^{3} \ln (3))$

Step 4.2.4.2

Add $12$ and $1$ .

$f (3) = \frac{1}{3} \cdot ({(\frac{13}{3})}^{3} \ln (3))$

Step 4.2.5

Apply the product rule to $\frac{13}{3}$ .

$f (3) = \frac{1}{3} \cdot (\frac{13^{3}}{3^{3}} \cdot \ln (3))$

Step 4.2.6

Raise $13$ to the power of $3$ .

$f (3) = \frac{1}{3} \cdot (\frac{2197}{3^{3}} \cdot \ln (3))$

Step 4.2.7

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

$f (3) = \frac{1}{3} \cdot (\frac{2197}{27} \cdot \ln (3))$

Step 4.2.8

Simplify $\frac{2197}{27} \ln (3)$ by moving $\frac{2197}{27}$ inside the logarithm.

$f (3) = \frac{1}{3} \cdot \ln (3^{\frac{2197}{27}})$

Step 4.2.9

Simplify $\frac{1}{3} \ln (3^{\frac{2197}{27}})$ by moving $\frac{1}{3}$ inside the logarithm.

$f (3) = \ln ({(3^{\frac{2197}{27}})}^{\frac{1}{3}})$

Step 4.2.10

Multiply the exponents in ${(3^{\frac{2197}{27}})}^{\frac{1}{3}}$ .

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Step 4.2.10.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (3) = \ln (3^{\frac{2197}{27} \cdot \frac{1}{3}})$

Step 4.2.10.2

Multiply $\frac{2197}{27} \cdot \frac{1}{3}$ .

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Step 4.2.10.2.1

Multiply $\frac{2197}{27}$ by $\frac{1}{3}$ .

$f (3) = \ln (3^{\frac{2197}{27 \cdot 3}})$

Step 4.2.10.2.2

Multiply $27$ by $3$ .

$f (3) = \ln (3^{\frac{2197}{81}})$

Step 4.2.11

The final answer is $\ln (3^{\frac{2197}{81}})$ .

$\ln (3^{\frac{2197}{81}})$

Step 4.3

Convert $\ln (3^{\frac{2197}{81}})$ to decimal.

$y = 29.79816294$

Step 5

The log function can be graphed using the vertical asymptote at $x = 0$ and the points $(1, 0), (2, 21.0543456), (3, 29.79816294)$ .

Vertical Asymptote: $x = 0$

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 21.054 \\ 3 & 29.798 \end{array}$

Step 6