Question

need help on the attachment

Answer 1

I got f'(x)=-(X^2+2x-8)/(x^2+8)^2 the continuous roots are x=-4, 2, discontinuity at x=+2(2)^(1/2),x=-2(2)^(1/2), so you don't count them. The relative max is x=2, is this right?

Answer 2

Okay, so take the first derivative and solve for zero, which you did. Now take the second derivative and apply the second derivative test.

Alternately, you could find out the max and min using the first derivative test.

Answer 3

You do this using the critical numbers that you have, which are -4 and 2.
Looks like you did it right to me.

Answer 4

f'x = (x^2+8) - (x-1)*2x
----------------- = 0 for stationary points
(x^2+8)^2

x^2 + 8 -2x^2 + 2x = 0
x^2 - 2x - 8 = 0
(x - 4)(x + 2) = 0
x = 4 or x= -2

check for nature of stationary points:

x = 3 f'x = 9 + 8 - 18 + 6 positive
x = 5 f'x = 25 + 8 - 50 + 10 - negative

so x = 4 gives local maximum

first option is correct

Answer 5

at x = -2

x = -3 f'x = 9 + 8 - 18 -4 negative
x = -1 f' positive

at x = -2 there is a local minimum

Answer 6

this is correct - i've confirmed it with wolframalpha

Answer 7

i used the first derivative test as finding second derivative looks a bit messy

Answer 8

Thanks!

Answer 9

no probs

Answer 10

x^2 + 8 -2x^2 + 2x = 0, you should get when you simplify the numerator is -x^2+2x+8 not x^2 - 2x - 8 = 0

Answer 11

i just multiplied -x^2+2x+8 = 0 by -1 to get x^2-2x-8 = 0
their roots are exactly the same 4 and -2

Answer 12

Thanks!