If f(3) = 14 and f '(x) ≥ 3 for 3 ≤ x ≤ 7, how small can f(7) possibly be?

Question

anonymous · Answer

@jim_thompson5910 hey jim u think u could help me

myininaya · Answer

Looks like we have that mean-value theorem here.

anonymous · Answer

can I show u what i did? idk where i messed up

myininaya · Answer

$$f'(c)=\frac{f(b)-f(a)}{b-a}$$
where c is between a and b.
(well they are using some number x is in between a and b; just replace c with x here if it makes more since to you)

myininaya · Answer

sure

anonymous · Answer

[f(7)-f(3)/(7-14)
[f(7)-14/-7
f(7)-14/-7>=3
c=-7

myininaya · Answer

oh no...
ok the change of x is 7-3 which is 4.
the change of y is f(7)-f(3) and since f(3)=14 we could say f(7)-14

myininaya · Answer

so we have that f'(c)>=3
and f'(c)=(f(b)-f(a))/(b-a)
so that means (f(b)-f(a))/(b-a)>=3.

myininaya · Answer

Your objective is to solve this inequality for f(7)

anonymous · Answer

ok sec let me try to work it out

anonymous · Answer

f(7)-14/4>=3

anonymous · Answer

f(7)=26?

myininaya · Answer

almost you had an inequality

myininaya · Answer

f(7)>=26

myininaya · Answer

This means f(7) is possibly 26 or bigger.

anonymous · Answer

yay thank you soo much for ur help!

myininaya · Answer

Np. You did most of the work. :)