Little help please

**mathlete** · January 25th 2008, 10:07 AM

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

**wingless** · January 25th 2008, 10:24 AM

$\frac{5x^2-7x+2}{4x^2-2}$

We can try to make the denominator 0.

$4x^2-2=0$

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

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

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

**mr fantastic** · January 25th 2008, 05:40 PM

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

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

Divide numerator and denominator by $x^2$ :

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

Therefore:

As $x \rightarrow \pm \infty$ , $f(x) \rightarrow \frac{5 - 0 + 0}{4 - 0} = \frac{5}{4}$ .

**JaneBennet** · January 25th 2008, 06:16 PM

Originally Posted by wingless

$\frac{5x^2-7x+2}{4x^2-2}$

We can try to make the denominator 0.

$4x^2-2=0$

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

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

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

You are making $\mathrm{f}(x)\to\pm\infty$ . The question is about $x\to\pm\infty$ .

**wingless** · January 26th 2008, 01:30 AM

Haha sorry I just misread it

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Little help please