Find the local maximum and minimum values of f using both the First and Second Derivative Tests. f(x)= x/(x^(2)+16). I got y=4 for local max and y=-4 for local minimum, but it says it's wrong. Not sure what I'm doing wrong.

Question

Find the local maximum and minimum values of f using both the First and Second Derivative Tests.  f(x)= x/(x^(2)+16).  I got y=4 for local max and y=-4 for local minimum, but it says it's wrong. Not sure what I'm doing wrong.

anonymous · Answer

first, write your f'(x)
hint: you'd have to se the quotient rule

anonymous · Answer

f'(x)= (x^(2)+16)*1-x*2x/(x^(2)+16)^(2) => x^(2)+16-2x^(2)/(x^(2)+16)^(2) =>16-x^(2)/(x^(2)+16)^(2) and I set x^(2)/(x^(2)+16)^(2) =0.

anonymous · Answer

you can use the equation editor button below so
people can understand what you just wrote :)

anonymous · Answer

$$f'(x)= (x^2+16)*1-x*2x/(x^2+16)^2$$

anonymous · Answer

$$x^2+16-2x^2/(x^2+16)^2 $$

anonymous · Answer

$$f'(x): 16-x^2/(x^2+16)^2=0$$

anonymous · Answer

so,
$$16-x^2=0\implies x=\pm4
$$

anonymous · Answer

one point should be the maxima and the other, the minima
we check using second derivative.
similarly, express your second derivative
f''(x)=?

anonymous · Answer

$$-2x^3-32x-64x+4x^3/(x^2+9)^3$$

anonymous · Answer

Refer to the attachment.  Calculations were performed by Mathematica.