Graph y=xe^x

$y = x e^{x}$

Find where the expression $x e^{x}$ 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.

The vertical asymptotes occur at areas of infinite discontinuity.

No Vertical Asymptotes

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

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

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

Apply L\'Hospital\'s rule.

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

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

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

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

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

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

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

Infinity divided by infinity is undefined.

Undefined

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

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.

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

Find the derivative of the numerator and denominator.

Differentiate the numerator and denominator.

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

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

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

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 $g (x) = - x$ .

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

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

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

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

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

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

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

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

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

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

Multiply $- 1$ by $1$ .

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

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

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

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

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

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

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

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

Move the negative in front of the fraction.

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

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

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

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

$- 0$

Multiply $- 1$ by $0$ .

$0$

List the horizontal asymptotes:

$y = 0$

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

This is the set of all asymptotes.

No Vertical Asymptotes

Horizontal Asymptotes: $y = 0$

No Oblique Asymptotes

Graph y=xe^x

Related Questions

Support LCMGCF.com by disabling your adblocker.