h = a (−2x+1)⁵ − b

A.

h' = a * 5(−2x+1)⁴ * −2 − 0

h' = −10a(−2x+1)⁴

h'' = −10a * 4(−2x+1)³ * −2

h'' = 80a(−2x+1)³

B.

h is monotonic if h' ≤ 0 for all x, or if h' ≥ 0

h' = −10a(−2x+1)⁴

For a < 0, −10a > 0 and (−2x+1)⁴ ≥ 0

Therefore, h' ≥ 0 for all x, and h is non-decreasing

For a > 0, −10a < 0 and (−2x+1)⁴ ≥ 0

Therefore, h' ≤ 0 for all x, and h is non-increasing

C.

Critical points occur where h' = 0

−10a(−2x+1)⁴ = 0, where a ≠ 0

(−2x+1)⁴ = 0

x = 1/2

The only critical point is at x = 1/2, which is independent of a or b.

I'll assume the 6 was supposed to be a b.

Use the chain rule, with u=-2x+1 to write:

h(x) = au^5 + b

dh/dx = dh/du * du/dx .... chain rule

h'(x) = 5au^4 * (-2) = -10au^4 = -10a(-2x + 1)^4

h"(x) = d(h')/du * du/dx = -40au^3 * (-2) = 80au^3 = 80a(-2x + 1)^3

For all x≠1/2, (-2x + 1) is nonzero and (-2x + 1)^4 is positive. The sign of h' is the same as the sign of -10a, always the same sign...opposite that of a. That makes h(x) monontonic. Increasing if a<0 or decreasing if a>0.

h(x) is defined and differentiable everywhere (it's a polynomial) so the only critical points are the zero-derivative type.

0 = h'(x) = -10a(-2x + 1)^4 clearly requires -2x+1 = 0, and x = 1/2 is the only critical point. That's independent of a and b.

h(x) = a(1-2x)⁵ + b

A.

h′(x) = 5a(1-2x)⁴(-2) = -10a(1-2x)⁴

h″(x) = -10a×4(1-2x)³(-2) = 80a(1-2x)³

B. h′(x) always remains negative or at the most becomes zero at x = 1/2, hence h(x) is monotonically decreasing.

C. For critical points, h′(x) = 0, x = 1/2 which is independent of a and b.

Now, 10 pts. pleeeeeeeeeeeeeeez!!!!

b is not used in the equation. Did you mean h(x) = a(-2x+1)⁵-b ?

h(x) = a(-2x+1)⁵-b

h'(x) = -10a(-2x+1)⁴

h"(x) = 80a(-2x+1)³

h' = -10a (1 - 2x)^4

h" = 80a (1 - 2x)^3