Find all values of c?

Question

Let f be the function given by f(x)=x^3-3x^2. Find all values of c that satisfy the conclusion of the Mean Value Theorem of differential calculus on the closed interval [0, 3].

 

I know that f'(x)=3x^2-6x and that f'(c)=3 but what to do next? Because the answer is 2. Please show all your work.

Answer 1

The Mean Value Theorem (MVT) states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that

f'(c)=(f(b)-f(a))/(b-a).

Your function f(x)=x³-3x² is continuous on [0,3] and differentiable on (0,3), so there exists at least one c in (0,3) such that

f'(c)=(f(3)-f(0))/(3-0)=(0-0)/3=0.

Since f'(x)=3x²-6x, we have f'(c)=3c²-6c=0, which has only one solution in (0,3), namely c=2 (since the other solution, c=0, is not in the open interval).

Answer 2

MVT: There is at LEAST one point whose tangent is parallel to the secant BETWEEN the two endpoints.

 

tangent: f'(x): This is the definition of the derivative of the function @ f(x)

Secant: This is defined as the slope between the two endpoints (a,f(a)), (b,f(b)) = (f(b)-f(a))/(b-a)

 

substituting (0,0) & (3,0) above = (0-0)/(3-0)=0.  Thus the secant slope is 0.

 

f'(x)= 3x^2 - 6x

 

Setting this equal to the secant of the two endpoints (Zero):

 

3x^2 - 6x = 0

 

x(3x - 6) = 0  yields two solutions

 

x=0 and x=2

 

The second solution (x=2) is between the two endpoints (X=0, X=3) and fulfills the requirement and thus verifies (for this example) that the MVT is true.