Find the absolute min and max of the given function on the interval [-4,1].

f(x)= -2x/((x^2+9)%^5)

I know you have to set the first derivative to 0 and solve for x to get the critical numbers. But I'm not sure if I simplified the equation right to get the right C.P.'s. The critical points I got were +-(3/2).

Question

Find the absolute min and max of the given function on the interval [-4,1]. 

f(x)= -2x/((x^2+9)%^5)


I know you have to set the first derivative to 0 and solve for x to get the critical numbers. But I'm not sure if I simplified the equation right to get the right C.P.'s. The critical points I got were +-(3/2).

saifoo.khan · Answer

What derivative did you get?

saifoo.khan · Answer

?

saifoo.khan · Answer

You should end up with:
x^2-1=0

anonymous · Answer

The derivative after my quotient rule is really long... How did you get that derivative? Did you simplify the equation?

anonymous · Answer

$$f \prime= (-2*(x ^{2}+9)^{(5/2)} + 2x*(5/2)*(x ^{2}+9)^{(3/2)}*2x) / (x ^{2}+9)^{5}$$

saifoo.khan · Answer

Give me 2 minutes. I'll show you.

anonymous · Answer

Thanks in advance!

saifoo.khan · Answer

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saifoo.khan · Answer

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anonymous · Answer

Oh my god. I wrote the function wrong. It's actually (x^2+9)^(5/2). I'm so sorry about the mistake.

saifoo.khan · Answer

Lol. Okay wait.

saifoo.khan · Answer

I'll brb in 15. Then i"ll solve.