1. ## Calculus Help Needed!

If f(2)=4 and f'(x)>2 for 2<x<11, how small can f(11) be?

Thanks!

2. Originally Posted by asnxbbyx113
If f(2)=4 and f'(x)>2 for 2<x<11, how small can f(11) be?

Thanks!
Note that f satisfies the conditions for the mean value theorem to be
applicable on [2,11].

Suppose f(11)<22, then by the mean value theorem there exists a c in (2,11)
such that:

f'(c)=(f(11)-f(2))/9 = (f(11)-4)/9,

but if f(11)<22 this is <2, which contradicts the assumption that f'(x)>=2.

Now let f(x)=4+2(x-2), then f'(x)=2 for x in (2,11), and f(11)=22.

Therefore f(11) can be as small as 22 and no smaller.

RonL

3. Hello, asnxbbyx113!

Did you make a sketch?

If $\displaystyle f(2)=4$ and $\displaystyle f'(x) \geq 2$ for $\displaystyle 2 \leq x \leq 11$, how small can $\displaystyle f(11)$ be?

We consider the graph of $\displaystyle f(x)$ from $\displaystyle x = 2$ to $\displaystyle x = 11$.

SInce $\displaystyle f(2) = 4$, the graph starts at $\displaystyle (2,4)$.
Code:
        | (2,4)
|   *
|
|
- + - + - - - - - - -
|   2

From $\displaystyle x = 2$ to $\displaystyle x = 11$, the slope is $\displaystyle \text{at least 2.}$

The very "lowest" graph would be straight line from $\displaystyle (2,4)$ with slope $\displaystyle 2$.
. . Its equation is: .$\displaystyle f(x) \:=\:2x$

It would look like this:
Code:
        |
|                 * (11,22)
|               * :
|             *   :
|           *     :
|         *       :
|       *         :
|     *           :
|   *             :
|   :             :
|   :             :
- | - + - - - - - - + - -
|   2            11

Hence, the least value for $\displaystyle f(11)$ is: .$\displaystyle f(11) = 22$

4. Originally Posted by CaptainBlank
Note that f satisfies the conditions for the mean value theorem to be
applicable on [2,11].
I am honored that I taught you something. I remember somebody asked a similar question. You answered it in some other way. I answered it elegantly via MVT. Now it seems you liked mine more.

5. Originally Posted by ThePerfectHacker
I am honored that I taught you something. I remember somebody asked a similar question. You answered it in some other way. I answered it elegantly via MVT. Now it seems you liked mine more.
Well I've always known about the use of the MVT, but I thought it
inappropriate for the level of knowlege of the student involved, as
it probably is here also, which is why Soroban is thought a better helper
than both of us .

A MVT proof is more appropriate to a first Real Analysis course than to
any course calling itself Calculus X.

RonL

6. Originally Posted by CaptainBlank
.