(x) -3 0 3 6 f(x) -5 4 1 7
f'(x) -1 2 -2 4
The table above gives values of a twice-differentiable function f and its first derivative f′ for selected values of x . Let g be the function defined by g(x)=f(x^2−x).
(a) What is the value of g′(3)?
(b) It is known that g′′(0)=−1 . What is the value of f′′(0)?
(c) Is there a value c , for 0 (d) Let h be the function with derivative given by h′(x)=4ecosx . At what value of x in the interval −3≤x≤0 does the instantaneous rate of change of h equal the average rate of change of f over the interval −3≤x≤0 ?

c is the correct answer

step-by-step explanation:

i would say b) is your answer

sorry if im wrong

step-by-step explanation:

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .

c

step-by-step explanation: