help me with this calculus problem please?

Question

let f be the function given by f(x)=x^3-7x+6
a) find the zeros fo f
b) write an equation of the line tangent to the graph of f at x=-1
c) find the number c that satisfies the conclusion of the Mean Value Theorem for f on the closed interval (1,3)

Answer 1

a) Synthetic division quickly gives you:
(x - 1)(x^2 + x - 6)
Then:
(x - 1)(x + 3)(x - 2)
So the zeros are:
x = 1, x = -3, x = 2

b) f(x) = x^3 -7x +6
dy/dx = 3x^2 - 7
slope at x = -1:
3(-1)^2 - 7 = 6 - 7 = -1
f(-1) = (-1)^3 -7(-1) + 6 = -1 + 7 + 6 = 12
The line that passes through the point (-1, 12) with slope -1:
y = -x + 11

c) There is a c in [1, 3] such that
f '(c) = [f(3) - f(1)]/(3 - 1) = (12 - 0)/2 = 6
Set the derivative equal to 6 and solve for x:
3x^2 - 7 = 6
3x^2 = 13
x^2 = 13/3
x = √(13/3)