Calculus Examples

$\tan (\ln (x))$

Step 1

Find the asymptotes.

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Step 1.1

For any $y = \tan (x)$ , vertical asymptotes occur at $x = \frac{π}{2} + n π$ , where $n$ is an integer. Use the basic period for $y = \tan (x)$ , $(- \frac{π}{2}, \frac{π}{2})$ , to find the vertical asymptotes for $y = \tan (\ln (x))$ . Set the inside of the tangent function, $b x + c$ , for $y = a \tan (b x + c) + d$ equal to $- \frac{π}{2}$ to find where the vertical asymptote occurs for $y = \tan (\ln (x))$ .

$x = - \frac{π}{2}$

Step 1.2

Set the inside of the tangent function $x$ equal to $\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 1.3

The basic period for $y = \tan (\ln (x))$ will occur at $(- \frac{π}{2}, \frac{π}{2})$ , where $- \frac{π}{2}$ and $\frac{π}{2}$ are vertical asymptotes.

$(- \frac{π}{2}, \frac{π}{2})$

Step 1.4

Find the period $\frac{π}{| b |}$ to find where the vertical asymptotes exist.

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Step 1.4.1

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 1.4.2

Divide $π$ by $1$ .

$π$

Step 1.5

The vertical asymptotes for $y = \tan (\ln (x))$ occur at $- \frac{π}{2}$ , $\frac{π}{2}$ , and every $π n$ , where $n$ is an integer.

$π n$

Step 1.6

There are only vertical asymptotes for tangent and cotangent functions.

Vertical Asymptotes: $x = \frac{π}{2} + π n$ for any integer $n$

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π}{2} + π n$ for any integer $n$

No Horizontal Asymptotes

No Oblique Asymptotes

Step 2

Find the point at $x = 1$ .

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Step 2.1

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

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

Step 2.2

Simplify the result.

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Step 2.2.1

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

$f (1) = \tan (0)$

Step 2.2.2

The exact value of $\tan (0)$ is $0$ .

$f (1) = 0$

Step 2.2.3

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.

$f (2) = \tan (\ln (2))$

Step 3.2

Simplify the result.

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Step 3.2.1

Evaluate $\tan (\ln (2))$ .

$f (2) = 0.83064087$

Step 3.2.2

The final answer is $0.83064087$ .

$0.83064087$

Step 4

Find the point at $x = 3$ .

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Step 4.1

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

$f (3) = \tan (\ln (3))$

Step 4.2

Simplify the result.

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Step 4.2.1

Evaluate $\tan (\ln (3))$ .

$f (3) = 1.95803332$

Step 4.2.2

The final answer is $1.95803332$ .

$1.95803332$

Step 5

The log function can be graphed using the vertical asymptote at $x = \frac{π}{2} + π n (f o r) (a n y) (i n t e g e r) n$ and the points $(1, 0), (2, 0.83064087), (3, 1.95803332)$ .

Vertical Asymptote: $x = \frac{π}{2} + π n (f o r) (a n y) (i n t e g e r) n$

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 0.831 \\ 3 & 1.958 \end{array}$

Step 6