Calculus Examples

$f (x) = - {(x - 0.086)}^{e^{- x}}$

Step 1

Find where the expression $- {(x - 0.086)}^{e^{- x}}$ is undefined.

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 2

The vertical asymptotes occur at areas of infinite discontinuity.

No Vertical Asymptotes

Step 3

Evaluate $lim_{x \to \infty} - {(x - 0.086)}^{e^{- x}}$ to find the horizontal asymptote.

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Step 3.1

Move the term $- 1$ outside of the limit because it is constant with respect to $x$ .

$- lim_{x \to \infty} {(x - 0.086)}^{e^{- x}}$

Step 3.2

Use the properties of logarithms to simplify the limit.

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Step 3.2.1

Rewrite ${(x - 0.086)}^{e^{- x}}$ as $e^{\ln ({(x - 0.086)}^{e^{- x}})}$ .

$- lim_{x \to \infty} e^{\ln ({(x - 0.086)}^{e^{- x}})}$

Step 3.2.2

Expand $\ln ({(x - 0.086)}^{e^{- x}})$ by moving $e^{- x}$ outside the logarithm.

$- lim_{x \to \infty} e^{e^{- x} \ln (x - 0.086)}$

Step 3.3

Move the limit into the exponent.

$- e^{lim_{x \to \infty} e^{- x} \ln (x - 0.086)}$

Step 3.4

Rewrite $e^{- x} \ln (x - 0.086)$ as $\frac{\ln (x - 0.086)}{e^{x}}$ .

$- e^{lim_{x \to \infty} \frac{\ln (x - 0.086)}{e^{x}}}$

Step 3.5

Apply L'Hospital's rule.

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Step 3.5.1

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

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Step 3.5.1.1

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

$- e^{\frac{lim_{x \to \infty} \ln (x - 0.086)}{lim_{x \to \infty} e^{x}}}$

Step 3.5.1.2

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

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

Step 3.5.1.3

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

$- e^{\frac{\infty}{\infty}}$

Step 3.5.1.4

Infinity divided by infinity is undefined.

Undefined

$- e^{\frac{\infty}{\infty}}$

Step 3.5.2

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.

$lim_{x \to \infty} \frac{\ln (x - 0.086)}{e^{x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (x - 0.086)]}{\frac{d}{d x} [e^{x}]}$

Step 3.5.3

Find the derivative of the numerator and denominator.

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Step 3.5.3.1

Differentiate the numerator and denominator.

$- e^{lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (x - 0.086)]}{\frac{d}{d x} [e^{x}]}}$

Step 3.5.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x - 0.086$ .

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Step 3.5.3.2.1

To apply the Chain Rule, set $u$ as $x - 0.086$ .

$- e^{lim_{x \to \infty} \frac{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x - 0.086]}{\frac{d}{d x} [e^{x}]}}$

Step 3.5.3.2.2

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

$- e^{lim_{x \to \infty} \frac{\frac{1}{u} \frac{d}{d x} [x - 0.086]}{\frac{d}{d x} [e^{x}]}}$

Step 3.5.3.2.3

Replace all occurrences of $u$ with $x - 0.086$ .

$- e^{lim_{x \to \infty} \frac{\frac{1}{x - 0.086} \frac{d}{d x} [x - 0.086]}{\frac{d}{d x} [e^{x}]}}$

Step 3.5.3.3

By the Sum Rule, the derivative of $x - 0.086$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 0.086]$ .

$- e^{lim_{x \to \infty} \frac{\frac{1}{x - 0.086} (\frac{d}{d x} [x] + \frac{d}{d x} [- 0.086])}{\frac{d}{d x} [e^{x}]}}$

Step 3.5.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$- e^{lim_{x \to \infty} \frac{\frac{1}{x - 0.086} (1 + \frac{d}{d x} [- 0.086])}{\frac{d}{d x} [e^{x}]}}$

Step 3.5.3.5

Since $- 0.086$ is constant with respect to $x$ , the derivative of $- 0.086$ with respect to $x$ is $0$ .

$- e^{lim_{x \to \infty} \frac{\frac{1}{x - 0.086} (1 + 0)}{\frac{d}{d x} [e^{x}]}}$

Step 3.5.3.6

Add $1$ and $0$ .

$- e^{lim_{x \to \infty} \frac{\frac{1}{x - 0.086} \cdot 1}{\frac{d}{d x} [e^{x}]}}$

Step 3.5.3.7

Multiply $\frac{1}{x - 0.086}$ by $1$ .

$- e^{lim_{x \to \infty} \frac{\frac{1}{x - 0.086}}{\frac{d}{d x} [e^{x}]}}$

Step 3.5.3.8

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

$- e^{lim_{x \to \infty} \frac{\frac{1}{x - 0.086}}{e^{x}}}$

Step 3.5.4

Multiply the numerator by the reciprocal of the denominator.

$- e^{lim_{x \to \infty} \frac{1}{x - 0.086} \cdot \frac{1}{e^{x}}}$

Step 3.5.5

Multiply $\frac{1}{x - 0.086}$ by $\frac{1}{e^{x}}$ .

$- e^{lim_{x \to \infty} \frac{1}{(x - 0.086) e^{x}}}$

Step 3.5.6

Reorder factors in $\frac{1}{(x - 0.086) e^{x}}$ .

$- e^{lim_{x \to \infty} \frac{1}{e^{x} (x - 0.086)}}$

Step 3.6

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{e^{x} (x - 0.086)}$ approaches $0$ .

$- e^{0}$

Step 3.7

Simplify the answer.

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Step 3.7.1

Anything raised to $0$ is $1$ .

$- 1 \cdot 1$

Step 3.7.2

Multiply $- 1$ by $1$ .

$- 1$

Step 4

List the horizontal asymptotes:

$y = - 1$

Step 5

There is no oblique asymptote because the degree of the numerator is less than or equal to the degree of the denominator.

No Oblique Asymptotes

Step 6

This is the set of all asymptotes.

No Vertical Asymptotes

Horizontal Asymptotes: $y = - 1$

No Oblique Asymptotes

Step 7