Finding f'(c) With an Inequality

Question

Suppose that f'(x)≤2 for 2≤x≤4. Show that f(4)-f(2)≤4.

Assume that f does satisfy the conditions of the Mean Value Thereom on the interval [2,4]. 

I'm using f(b)-f(a)/b-a=f'(c). So I'm assuming a=2 and b=4. And I'm also pretty sure that 2≤c≤4. But now I need to solve for f(4)-f(2) but I don't know what f is! 

Answer 1

You don't know what f is, but you know the maximum amount that it could increase over the interval. You have

 

[f(b)-f(a)]/[b-a] = f'(c)

 

and you know that f'(x)≤2, so

 

[f(b)-f(a)]/[b-a] ≤ 2

 

Plug in a=2 and b=4 and simplify to get

[f(4) - f(2)] / [4-2] ≤ 2

[f(4) - f(2)] / [2] ≤ 2

f(4) - f(2) ≤ 4