Calculus Examples

$f (x) = (8 - 6 x) e^{x}$

Step 1

Find where the expression $8 e^{x} - 6 x e^{x}$ 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 $lim_{x \to - \infty} 8 e^{x} - 6 x e^{x}$ to find the horizontal asymptote.

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Step 3.1

Evaluate the limit.

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Step 3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- \infty$ .

$lim_{x \to - \infty} 8 e^{x} - lim_{x \to - \infty} 6 x e^{x}$

Step 3.1.2

Move the term $8$ outside of the limit because it is constant with respect to $x$ .

$8 lim_{x \to - \infty} e^{x} - lim_{x \to - \infty} 6 x e^{x}$

Step 3.2

Since the exponent $x$ approaches $- \infty$ , the quantity $e^{x}$ approaches $0$ .

$8 \cdot 0 - lim_{x \to - \infty} 6 x e^{x}$

Step 3.3

Move the term $6$ outside of the limit because it is constant with respect to $x$ .

$8 \cdot 0 - 6 lim_{x \to - \infty} x e^{x}$

Step 3.4

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

$8 \cdot 0 - 6 lim_{x \to - \infty} \frac{x}{e^{- x}}$

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.

$8 \cdot 0 - 6 \frac{lim_{x \to - \infty} x}{lim_{x \to - \infty} e^{- x}}$

Step 3.5.1.2

The limit at negative infinity of a polynomial of odd degree whose leading coefficient is positive is negative infinity.

$8 \cdot 0 - 6 \frac{- \infty}{lim_{x \to - \infty} e^{- x}}$

Step 3.5.1.3

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

$8 \cdot 0 - 6 \frac{- \infty}{\infty}$

Step 3.5.1.4

Infinity divided by infinity is undefined.

Undefined

$8 \cdot 0 - 6 \frac{- \infty}{\infty}$

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.

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

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.

$8 \cdot 0 - 6 lim_{x \to - \infty} \frac{\frac{d}{d x} [x]}{\frac{d}{d x} [e^{- x}]}$

Step 3.5.3.2

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

$8 \cdot 0 - 6 lim_{x \to - \infty} \frac{1}{\frac{d}{d x} [e^{- x}]}$

Step 3.5.3.3

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

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Step 3.5.3.3.1

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

$8 \cdot 0 - 6 lim_{x \to - \infty} \frac{1}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x]}$

Step 3.5.3.3.2

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

$8 \cdot 0 - 6 lim_{x \to - \infty} \frac{1}{e^{u} \frac{d}{d x} [- x]}$

Step 3.5.3.3.3

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

$8 \cdot 0 - 6 lim_{x \to - \infty} \frac{1}{e^{- x} \frac{d}{d x} [- x]}$

Step 3.5.3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$8 \cdot 0 - 6 lim_{x \to - \infty} \frac{1}{e^{- x} \cdot - 1 \frac{d}{d x} [x]}$

Step 3.5.3.5

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

$8 \cdot 0 - 6 lim_{x \to - \infty} \frac{1}{e^{- x} \cdot - 1 \cdot 1}$

Step 3.5.3.6

Multiply $- 1$ by $1$ .

$8 \cdot 0 - 6 lim_{x \to - \infty} \frac{1}{e^{- x} \cdot - 1}$

Step 3.5.3.7

Move $- 1$ to the left of $e^{- x}$ .

$8 \cdot 0 - 6 lim_{x \to - \infty} \frac{1}{- 1 \cdot e^{- x}}$

Step 3.5.3.8

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$8 \cdot 0 - 6 lim_{x \to - \infty} \frac{1}{- e^{- x}}$

Step 3.5.4

Cancel the common factor of $1$ and $- 1$ .

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Step 3.5.4.1

Rewrite $1$ as $- 1 (- 1)$ .

$8 \cdot 0 - 6 lim_{x \to - \infty} \frac{- 1 (- 1)}{- e^{- x}}$

Step 3.5.4.2

Move the negative in front of the fraction.

$8 \cdot 0 - 6 lim_{x \to - \infty} - \frac{1}{e^{- x}}$

Step 3.6

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

$8 \cdot 0 - 6 \cdot - 1 lim_{x \to - \infty} \frac{1}{e^{- x}}$

Step 3.7

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

$8 \cdot 0 - 6 \cdot - 1 \cdot 0$

Step 3.8

Simplify terms.

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Step 3.8.1

Multiply $- 1$ by $0$ .

$8 \cdot 0 - 6 \cdot 0$

Step 3.8.2

Simplify the answer.

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Step 3.8.2.1

Simplify each term.

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Step 3.8.2.1.1

Multiply $8$ by $0$ .

$0 - 6 \cdot 0$

Step 3.8.2.1.2

Multiply $- 6$ by $0$ .

$0 + 0$

Step 3.8.2.2

Add $0$ and $0$ .

$0$

Step 4

List the horizontal asymptotes:

$y = 0$

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: $y = 0$

No Oblique Asymptotes

Step 7