• Jan 25th 2008, 10:07 AM
mathlete
f(x)=(5x^2-7x+2)/(4x^2-2)

f(x)--> ? as x--> -infinity

f(x)--> ? as x--> infinty

can someone help me out here, i seem to be coming up with the wrong answers..thanks
• Jan 25th 2008, 10:24 AM
wingless
$\displaystyle \frac{5x^2-7x+2}{4x^2-2}$

We can try to make the denominator 0.

$\displaystyle 4x^2-2=0$

$\displaystyle x^2 = \frac{1}{2}$

$\displaystyle x = \{ \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} \}$

Now, plug those values in the function and check which one makes it $\displaystyle +\infty$ or $\displaystyle - \infty$
• Jan 25th 2008, 05:40 PM
mr fantastic
Quote:

Originally Posted by mathlete
f(x)=(5x^2-7x+2)/(4x^2-2)

f(x)--> ? as x--> -infinity

f(x)--> ? as x--> infinty

can someone help me out here, i seem to be coming up with the wrong answers..thanks

$\displaystyle f(x) = \frac{5x^2 - 7x + 2}{4x^2 - 2}$.

Divide numerator and denominator by $\displaystyle x^2$:

$\displaystyle f(x) = \frac{5 - \frac{7}{x} + \frac{2}{x^2}}{4 - \frac{2}{x^2}}$.

Therefore:

As $\displaystyle x \rightarrow \pm \infty$, $\displaystyle f(x) \rightarrow \frac{5 - 0 + 0}{4 - 0} = \frac{5}{4}$.
• Jan 25th 2008, 06:16 PM
JaneBennet
Quote:

Originally Posted by wingless
$\displaystyle \frac{5x^2-7x+2}{4x^2-2}$

We can try to make the denominator 0.

$\displaystyle 4x^2-2=0$

$\displaystyle x^2 = \frac{1}{2}$

$\displaystyle x = \{ \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} \}$

Now, plug those values in the function and check which one makes it $\displaystyle +\infty$ or $\displaystyle - \infty$

You are making $\displaystyle \mathrm{f}(x)\to\pm\infty$. The question is about $\displaystyle x\to\pm\infty$.
• Jan 26th 2008, 01:30 AM
wingless
Haha sorry I just misread it :rolleyes: