Question

Suppose that 3≤f'(x)≤5 for all values of x. show that 18<≤ f(8)-f(2)≤30

Answer 1

Do you know the Mean Value Theorem?

Answer 2

If f is a function defined on [a,b], f continuous on [a,b] and f is differntiable on (a,b), then there exists a point c in (a,b) such that:\[f(b)-f(a)=f^{\prime}(c)(b-a)\]

Answer 3

If you use this, with your interval being [2,8], you will get the desired result.

Answer 4

Notice that 8-2=6, and that 18=6*3, and 30=6*5.

Answer 5

I know the mean value theorem but i still dont understand the question and the way you described it.

Answer 6

Youre given that:\[3\le f^{\prime}(x) \le 5\]This is a fact about the derivative of f. You want to show:\[18 \le f(8)-f(2)\le 30\]This is something about the function f itself. The Mean Value Theorem is a great tool for relating values of the function to the derivative of the function.

Answer 7

If you plug in a=2 and b=8 into the MVT, it should be clear how to proceed.

Answer 8

ohh I got it. Thankyou