Question

Suppose f(x) is continuous on [3,8] and −2≤f′(x)≤3 for all x in (3,8). Use the Mean Value Theorem to estimate f(8)−f(3).

Answer 1

MVT: \(\large f'(c)=\frac{f(8)-f(3)}{8-3} \) for some c in (3,8)

Answer 2

but you can't evaluate the top? ? there's no f(x) given?

Answer 3

no... the top is what you need... maybe rearrange it so you'll have: \(\large f(8)-f(3)=f'(c)\cdot (8-3) \)

Answer 4

it's making no sense to me. Can you go step by step please?

Answer 5

any ideas, @Zarkon ???

Answer 6

not sure how you could be more clear

Answer 7

for any c in (3, 8), the minimum f' value is -2 and the maximum f' value is 3

Answer 8

i'm thinking the estimate for f(8)-f(3) will have to be between -2*5 and 3*5 ???
i dunno what else we can take from the given info... :(

Answer 9

;/ I'll go read over the chapter again. Thank you for your help.. (trying) I appreciate it.

Answer 10

sorry... i'll review my MVT also... :)