Suppose that 4 ≤ f '(x) ≤ 5 for all values of x. What are the minimum and maximum possible values of f(3) − f(1)?

Question

≤ f(3) − f(1) ≤

Please work out

Answer 1

By the Mean Value Theorem (which applies since f' is defined for all x), we know that

f(3) = f(1) + f'(x*)*(3-1), where 1 <= x* <= 3.

Thus,

f(3) - f(1) = 2f'(x*). This shows that the greatest possible value of f(3) - f(1) is 2*5 = 10, since 5 is an upper bound for the derivative f'.

Likewise, a lower bound for the difference f(3) - f(1) is 2*4 = 8.

Answer 2

By the mean value theorem, there exists c in the range [1,3] such that:

(f(3) - f(1)) / (3 - 1) = f'(c)

f(3) - f(1) = 2f'(c)

Since f'(c) is between 4 and 5, the above equation says that f(3) - f(1) must have a minimum of 8 and a maximum of 10.