# Suppose that 2 ≤ f '(x) ≤ 5 for all values of x. What are the minimum and maximum possible values of f(6) − f(2)? ? ≤ f(6) − f(2) ≤ ?

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### 1 Answer

If `2 <= f'(x) <= 5` then `2 <= (f(6)-f(2))/(6-2) <= 5`

**Then `8 <= f(6)-f(2) <= 20` **

** The function can never grow at a faster rate than 5; growing at the constant maximum rate of 5 on the interval from 2 to 6 means the value of the function increases by 5*4=20 units.

The function can never grow slower than a rate of 2. If the function grows at the constant minimum rate of 2 on the interval from 2 to 6 then the function increases by 4*2=8 units. **

This is similar to the mean value theorem type problems.