How can I find the max/min of this function?

[randy17]

Homework Statement

Find the critical point of f(x)=(-5x^2+3x)/(2x^2-5), and show whether this point is a max/min.

Homework 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?

 

[clamtrox]

What do I do now?

Try again. Everything looks OK.

 

[SteamKing]

Make sure you are using the right quadratic formula. Remember, it starts out with '-b', and your answers don't have the correct value for '-b'.

 

[MarneMath]

The op's quadratic answer is correct.* If you're plugging in the exact formula into the calculator, I would simply instead just take an approximation of your answers (ie .311 and 8.02) and plug that in and then proceed with determining if the values are a max or a min.

*(The -b is negated by the 2a, where a is negative.)

 

[Ray Vickson]

Homework Statement

Find the critical point of f(x)=(-5x^2+3x)/(2x^2-5), and show whether this point is a max/min.

Homework 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?

Your answers are correct; it sounds like you need to buy a new calculator.