Graph f(x)=1/(1+e^(-x))

$f (x) = \frac{1}{1 + e^{- x}}$

Step 1

Find where the expression $\frac{1}{1 + e^{- x}}$ 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 $lim_{x \to \infty} \frac{1}{1 + e^{- x}}$ to find the horizontal asymptote.

Evaluate the limit.

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $\infty$ .

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

Evaluate the limit of $1$ which is constant as $x$ approaches $\infty$ .

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

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

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

Evaluate the limit of $1$ which is constant as $x$ approaches $\infty$ .

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

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

$\frac{1}{1 + 0}$

Simplify the answer.

Add $1$ and $0$ .

$\frac{1}{1}$

Divide $1$ by $1$ .

$1$

Step 4

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

$0$

Step 5

List the horizontal asymptotes:

$y = 1, 0$

Step 6

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 7

This is the set of all asymptotes.

No Vertical Asymptotes

Horizontal Asymptotes: $y = 1, 0$

No Oblique Asymptotes

Step 8

Graph f(x)=1/(1+e^(-x))

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