plz help i keep on getting this not right?

Question

http://i181.photobucket.com/albums/x85/ctti_3/untitled-17.jpg

Answer 1

f(x) = 2(x-1/16x^2)
at -infinity, this is very negative.
at -1 this is about -2

Differentiate
f'(x) = 2 - (2/16 * -2 * x^-3)
Set this equal to zero, since at a maximum the slope will be zero. Multiply by x^3.
So 0 = 2x^3 + 1/16
So x^3 = -1/32
But then x > -1 so, the maximum is at -1
.

Answer 2

f(x) = 2x - (1/8)x^-2
f'(x) = 2 - (-2)(1/8)x^-3
f"(x) = -(-3)(-2)(1/8)x^-4

To find a critical point (maximum, minimum, inflection), make f''(x) = 0

To analyse a critical point, find its second derivative (f"): if it is positive, you have found a minimum; if negative, it is a max; if zero, most probably an inflection point -- check the neighbourhood.

f'(x) = 2 + (1/4)x^-3
we seek:
0 = 2 + (1/4)x^-3
-2 = (1/4)x^-3
-8 = x^-3
- 1/8 = x^3
- 1/2 = x

At x= -1/2, the second derivative is negative (because x^4 is always positive and the factor is negative)

This means that the slope (f') goes from positive to negative, meaning that f(-1/2) is a maximum.

If you go 'backwards' from x=-1/2 all the way back to - infinity, then the value of f(x) will go down all the time (otherwise, we would have found one or more critical points below x=-1 -- and we did not)

Therefore, on the interval, the maximum must be at x = -1.

Answer 3

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