Because we will have to do it eventually, let's find f(x) first.
f''(x) = 24x - 18
First, we integrate f''(x) using an indefinite integral.
f'(x) = ∫ (24x - 18) dx
f'(x) =Â 12X^2 - 18x + C
Now, we need to find C by substituting "x" for -1 and setting the equation equal to -6 because f'(-1) = -6
f'(-1) = 12(-1)^2 - 18(-1) + C = -6
Solve for C
C=-26
Now we put that into f'(x).
f'(x) = 12x^2 - 18x -26
Now, we integrate again.
f(x) = ∫(12x^2 - 18x - 26)dx
f(x) = 4x^3 - 9x^2 - 26x + C
Now, we substitute "x" for 2 and set it equal to 0.
f(2) = 4(2)^3 - 9(2)^2 - 26(2) + C = 0
Solve for C
C=38
Now that we have f(x), B has been solved.
Next, we need to find out where the slope of f(x) is equal to 0. Â Remember that to find slope, we need to find the derivative. Â We already found the derivative of f(x), so we can use that. Â The question asks for the places where the slope of f(x) is 0, so we need to set f'(x) equal to 0 and solve for "x"
12x^2 - 18x -26 = 0
I could try to factor this, but I know that it is not possible. Â We must use the quadratic formula. Â I cannot reasonable put the quadratic formula into this, so I will only do the part under the radical.
√[18^2-4(12)(-26)]
√(324+1248)
√1572
√(4)*√(393)
2√393
When this is done with the rest of the formula, you get.
[18+/-2√(393)]/24
Thus, the points where the slope of f(x) is equal to zero are where [18+/-2√(393)]/24 are the x values of f(x).
(  [18+/-2√(393)]/24 , f ( [18+/-2√(393)]/24 )  )
Now, we have done A and B.
To do C, we must remember the formula for finding the average value of f(x)
This formula is:
[The definite integral of f(x)] / (b-a)
The picture is of this formula.
When we solve for this, we get -13.
I hope I got everything right.
