# Function analysis

• October 24th 2009, 06:20 PM
live_laugh_luv27
Function analysis
$
f(x)=\frac{x^2+3x-4}{x^2+x-2}
$

a. Find the domain of function

b. Find $\lim_{x\to1}f(x)$

c. Find all vertical and horizontal asymptotes

d. Find $\lim_{x\to\infty}f(x)$

Here are my answers...are they correct?

a. All real numbers not equal to -2

b. $\infty$

c. Vertical: $y=1, -1$
Horizontal: $x=2$

d. $\lim_{x\to\infty}f(x)=1$
• October 24th 2009, 06:43 PM
Prove It
Quote:

Originally Posted by live_laugh_luv27
$
f(x)=\frac{x^2+3x-4}{x^2+x-2}
$

a. Find the domain of function

b. Find $\lim_{x\to1}f(x)$

c. Find all vertical and horizontal asymptotes

d. Find $\lim_{x\to\infty}f(x)$

Here are my answers...are they correct?

a. All real numbers not equal to -2

b. $\infty$

c. Vertical: $y=1, -1$
Horizontal: $x=2$

d. $\lim_{x\to\infty}f(x)=1$

$f(x)=\frac{x^2+3x-4}{x^2+x-2}$

a) Only half right.

Polynomials are continuous everywhere, so the quotient of polynomials is continuous everywhere except where the denominator is $0$.

So if the denominator is $0$, we have

$x^2 + x - 2 = 0$

$(x + 2)(x - 1) = 0$

$x = -2$ or $x = 1$.

Thus the domain of the function is $\mathbf{R}\backslash\{-2, 1\}$.

b) Incorrect.

Notice that we can rewrite the function as

$f(x) = \frac{(x + 4)(x - 1)}{(x + 2)(x - 1)}$

$= \frac{x + 4}{x + 2}$.

Now find $\lim_{x \to 1}f(x)$.

c) Nearly correct.

We can further rewrite the function as

$f(x) = 1 + \frac{2}{x + 2}$

There will be a vertical asymptote at $x = -2$ and a horizontal asymptote at $f(x) = 1$.

d) Correct.

As $x \to \infty, x + 2 \to \infty$.

Therefore $\frac{2}{x + 2} \to 0$.

Thus $1 + \frac{2}{x + 2} \to 1$.
• October 24th 2009, 06:47 PM
live_laugh_luv27
So for b, would the correct answer be -2? I'm still unsure of this part.
• October 24th 2009, 06:48 PM
Prove It
Quote:

Originally Posted by live_laugh_luv27
So for b, would the correct answer be -2? I'm still unsure of this part.

No.

$\lim_{x \to 1}\frac{x + 4}{x + 2} = \frac{1 + 4}{1 + 2}$

$= \frac{5}{3}$.