Question

Suppose f is a differentiable function such that f'(x)<=3 for all x E(2,7). if f(-2)=3 the Mean Value Theorem says that f(7)<=V for what value of V? (Choose V as small as possible)

Answer 1

is it \((2,7)\) or \((-2,7)\) ?

Answer 2

you have as an interval \((2,7)\) but then you have \(f(-2)\) so there is some confusion here

Answer 3

its -2 sorry lol

Answer 4

ok
how far from \(-2\) to \(7\) ?

Answer 5

5?

Answer 6

oh no

Answer 7

9

Answer 8

whew
ok so imagine you are walking, but you cannot walk faster than 3 mph
you walk for 9 hours
what is the furthest you can go?

Answer 9

27

Answer 10

right
and you start at mile maker \(3\)
so if you go at most 27 miles, what mile marker could you reach?

Answer 11

would it be 30 ?

Answer 12

yes
and that is one way to interpret the problem, and probably do it in your head
we can do it the other way if you like

Answer 13

is there an easier way lol

Answer 14

how did u know i was walking 3 mph

Answer 15

and how did u know marker was set at 3

Answer 16

\[f'(x)<3\] is like saying your speed is at most 3 mph
\(f(-2)=3\) means when you start counting you are at mile marker 3, and since you cannot walk more that 27 miles in 9 hours, you can be at most at mile marker 30
now lets do it the calc way

Answer 17

okay

Answer 18

by the mvthm
\[\frac{f(7)-f(-2)}{7-(-2)}=f'(c)\] for some \(c\)
but since \(f'(x)<3\) we have
\[\frac{f(7)-f(-2)}{7-(-2)}\leq 3\]

Answer 19

you know \(f(-2)=3\) and so this is the same as
\[\frac{f(7)-3)}{9}\leq 3\] or
\[f(7)-3\leq 27\] or
\[f(7)\leq 30\]

Answer 20

oh how cool this is shorter in a way

Answer 21

is this Rolls formula?