Consider the following function and closed interval. f(x) = (x − 4)(x − 6)(x − 7), [4, 7]

Question

Consider the following function and closed interval.

f(x) = (x − 4)(x − 6)(x − 7),    [4, 7]

If f is differentiable on the open interval (4, 7), find f '(x).

 (If it is not differentiable on the open interval, enter DNE.)

f '(x) =  

If Rolle's Theorem can be applied, find all values of c in the open interval (a, b)

 such that f '(c) = 0.

 (Enter your answers as a comma-separated list. If Rolle's Theorem cannot be applied, enter NA.)

c=

Answer 1

The given function is a polynomial, which means we can assume the function is differentiable everywhere:

f'(x) = (x-4)'(x-6)(x-7) + (x-4)(x-6)'(x-7) + (x-4)(x-6)(x-7)' = (x-6)(x-7) + (x-4)(x-7) + (x-4)(x-6) = (x²-13x+42) + (x²-11x+28) + (x²-10x+24) = 3x²-34x+94

One of the requirements for Rolle's Theorem is that f(a)=f(b), so f(4)=(4-4)(4-6)(4-7)=0, and f(7)=(7-4)(7-6)(7-7)=0, so this requirement is met. The other requirements were met previously because the function is continuous on the closed interval and differentiable on the open interval, so Rolle's Theorem can be applied!

Hence, now we find all values of c in the open interval (a,b) such that f'(c)=0:

f'(c) = 3c²-34c+94

0 = 3c²-34c+94

c = (-(-34)±√((-34)²-4(3)(94)))/(2(3))

c = (34±√(1156-1128))/6

c = (34±√28)/6

c = (34±2√7)/6

c = (17±√7)/3

Thus, the values of c are (17+√7)/3≈6.549 and (17-√7)/3≈4.785

Hope this helped!

Answer 2