How do I find the max min of this function?

[randy17]

The problem statement, all variables and given/known data
Find the critical point of f(x)=(-5x^2+3x)/(2x^2-5), and show whether this point is a max/min.

Relevant equations

f(x)=(-5x^2+3x)/(2x^2-5)

The attempt at a solution
When I tried solving for the derivative of; f(x)=(-5x^2+3x)/(2x^2-5), I got (-6x^2+50x-15)/2x^2-5)^2.

then I set it equal to 0 and I got 0=-6x^2+50x-15

Now from here do I use the quadratic formula to solve for x? I tried that and I am getting x=(25+sqrt535)/6 and x=(25-sqrt535)/6. Have I done it right so far? When I try to plug in the x-values into the original eq to get y my calc says error... What do I do now?

[mathmari]

Yes, you has done it right...

For [latex] x<=(25-\sqrt{535})/6 [/latex], [latex]f'(x)<=0 [/latex]
For [latex] (25-\sqrt{535})/6<x<(25+\sqrt{535})/6 [/latex], [latex]f'(x)>0 [/latex]
For [latex] x>=(25+\sqrt{535})/6 [/latex], [latex] f'(x)<=0 [/latex]

(The domain of the function is [latex]R[/latex] \ { [latex]-\sqrt{\frac{5}{2}}, \sqrt{\frac{5}{2}} [/latex] } )

So the min of this function is at [latex] x=(25-\sqrt{535})/6 [/latex], and it's[latex] f((25-\sqrt{535})/6)=(\sqrt{535}-25)/20[/latex]
And the max is at [latex] x=(25+\sqrt{535})/6[/latex], [latex] f((25+\sqrt{535})/6)=-(\sqrt{535}+25)/20[/latex]

Edited June 3, 2013 by mathmari

[daniton]

Why don't you just second derivative test.It's easy and quick.