Finite Math Examples

$y = e^{- x} \cdot \ln (x)$

Step 1

Find the asymptotes.

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Step 1.1

Find where the expression $e^{- x} \cdot \ln (x)$ is undefined.

$x \leq 0$

Step 1.2

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

$x = 0$

Step 1.3

Evaluate $lim_{x \to \infty} e^{- x} \ln (x)$ to find the horizontal asymptote.

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Step 1.3.1

Rewrite $e^{- x} \ln (x)$ as $\frac{\ln (x)}{e^{x}}$ .

$lim_{x \to \infty} \frac{\ln (x)}{e^{x}}$

Step 1.3.2

Apply L'Hospital's rule.

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Step 1.3.2.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 1.3.2.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{x \to \infty} \ln (x)}{lim_{x \to \infty} e^{x}}$

Step 1.3.2.1.2

As log approaches infinity, the value goes to $\infty$ .

$\frac{\infty}{lim_{x \to \infty} e^{x}}$

Step 1.3.2.1.3

Since the exponent $x$ approaches $\infty$ , the quantity $e^{x}$ approaches $\infty$ .

$\frac{\infty}{\infty}$

Step 1.3.2.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{\infty}{\infty}$

Step 1.3.2.2

Since $\frac{\infty}{\infty}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to \infty} \frac{\ln (x)}{e^{x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (x)]}{\frac{d}{d x} [e^{x}]}$

Step 1.3.2.3

Find the derivative of the numerator and denominator.

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Step 1.3.2.3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (x)]}{\frac{d}{d x} [e^{x}]}$

Step 1.3.2.3.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{d}{d x} [e^{x}]}$

Step 1.3.2.3.3

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

$lim_{x \to \infty} \frac{\frac{1}{x}}{e^{x}}$

Step 1.3.2.4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to \infty} \frac{1}{x} \cdot \frac{1}{e^{x}}$

Step 1.3.2.5

Multiply $\frac{1}{x}$ by $\frac{1}{e^{x}}$ .

$lim_{x \to \infty} \frac{1}{x e^{x}}$

Step 1.3.3

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x e^{x}}$ approaches $0$ .

$0$

Step 1.4

List the horizontal asymptotes:

$y = 0$

Step 1.5

No oblique asymptotes are present for logarithmic and trigonometric functions.

No Oblique Asymptotes

Step 1.6

This is the set of all asymptotes.

Vertical Asymptotes: $x = 0$

Horizontal Asymptotes: $y = 0$

Vertical Asymptotes: $x = 0$

Horizontal Asymptotes: $y = 0$

Step 2

Find the point at $x = 1$ .

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Step 2.1

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

$f (1) = e^{- (1)} \cdot \ln (1)$

Step 2.2

Simplify the result.

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Step 2.2.1

Multiply $- 1$ by $1$ .

$f (1) = e^{- 1} \cdot \ln (1)$

Step 2.2.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

$f (1) = 0$

Step 2.2.5

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.

$f (2) = e^{- (2)} \cdot \ln (2)$

Step 3.2

Simplify the result.

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Step 3.2.1

Multiply $- 1$ by $2$ .

$f (2) = e^{- 2} \cdot \ln (2)$

Step 3.2.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 3.2.3

Combine $\frac{1}{e^{2}}$ and $\ln (2)$ .

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

Step 3.2.4

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

$\frac{\ln (2)}{e^{2}}$

Step 3.3

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

$y = 0.09380727$

Step 4

Find the point at $x = 3$ .

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Step 4.1

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

$f (3) = e^{- (3)} \cdot \ln (3)$

Step 4.2

Simplify the result.

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Step 4.2.1

Multiply $- 1$ by $3$ .

$f (3) = e^{- 3} \cdot \ln (3)$

Step 4.2.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 4.2.3

Combine $\frac{1}{e^{3}}$ and $\ln (3)$ .

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

Step 4.2.4

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

$\frac{\ln (3)}{e^{3}}$

Step 4.3

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

$y = 0.05469668$

Step 5

The log function can be graphed using the vertical asymptote at $x = 0$ and the points $(1, 0), (2, 0.09380727), (3, 0.05469668)$ .

Vertical Asymptote: $x = 0$

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 0.094 \\ 3 & 0.055 \end{array}$

Step 6