Given that f(1) = 3 and 2 less than or equal to f'(x) less than or equal to 3 for all x, what is the smallest possible value for f(4)?

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- Jul 27th 2006, 12:21 PMNimmyCalculus I Homework Help part II
Given that f(1) = 3 and 2 less than or equal to f'(x) less than or equal to 3 for all x, what is the smallest possible value for f(4)?

- Jul 27th 2006, 12:25 PMCaptainBlackQuote:

Originally Posted by**Nimmy**

f(1)+3*2=9

RonL - Jul 27th 2006, 12:57 PMThePerfectHackerQuote:

Originally Posted by**Nimmy**

Honestly, I do not understand what Captain**Blank**is doing.

You are going to use the most impostant theorem in calculus,*mean-value theorem*.

To remind you.

If a function $\displaystyle f$ is countinous on $\displaystyle [a,b]$ and diffrenciable on $\displaystyle (a,b)$ then there exists are number $\displaystyle a<c<b$ such as,

$\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}$

-----

Now comes the answer.

Since $\displaystyle 2\leq f'(x)\leq 3$ for all $\displaystyle x$ you know that the function is diffrenciable for any $\displaystyle x$ (because it exists between 2 and 3). Furthermore, the function is countinous for any $\displaystyle x$ because diffrenciability implies counintuity.

Consider $\displaystyle f(x)$ on the interval $\displaystyle [1,4]$. By the previous explanation we have shown this function satisfies the conditions of the mean value theorem. Thus there is a $\displaystyle 1<c<4$ such as,

$\displaystyle f'(c)=\frac{f(4)-f(1)}{4-1}$. Thus,

Thus,

$\displaystyle 3f'(c)=f(4)-f(1)$

Thus,

$\displaystyle 3f'(c)=f(4)-3$

Thus,

$\displaystyle f(4)=3f'(c)+3$

Therefore the largest possible value of $\displaystyle f(4)$ is the largest possible value of $\displaystyle f'(c)$ which is 3. Thus,

$\displaystyle \max\{f(4)\}=3(3)+3=12$

And the smallest possible values is similarly,

$\displaystyle \min{f(4)\}=3(2)+3=9$ - Jul 27th 2006, 01:44 PMtopsquarkQuote:

Originally Posted by**ThePerfectHacker**

**Black**'s method was much simpler. I had to read yours twice to see what you were doing. ;)

-Dan - Jul 27th 2006, 01:59 PMThePerfectHackerQuote:

Originally Posted by**topsquark**

- Jul 27th 2006, 02:06 PMtopsquarkQuote:

Originally Posted by**ThePerfectHacker**

You basically said the same thing, just with a lot more precision.

-Dan

Tsk, tsk! You aren't supposed to edit quotes! :p - Jul 27th 2006, 07:47 PMSoroban
Hello, Nimmy!

Quote:

Given that $\displaystyle f(1) = 3$ and $\displaystyle 2 \leq f '(x) \leq 3$ for all $\displaystyle x$,

what is the smallest possible value for $\displaystyle f(4)$ ?

This is what the Captain meant . . .Code:`| Q`

| * (4,?)

| :

| :

| P :

| * :

| (1,3) :

| :

| :

- - + - + - + - + - + - -

| 1 2 3 4

We are given point $\displaystyle P(1,3)$

When $\displaystyle x = 4$, we have a point $\displaystyle Q(4,?)$

$\displaystyle 2 \leq f'(x) \leq 3$ means:

. . the slope of the graph from $\displaystyle P$ to $\displaystyle Q$ is between $\displaystyle 2$ and $\displaystyle 3$.

The*lowest*$\displaystyle Q$ occurs when the slope is**exactly 2**(all the way).

This puts $\displaystyle Q$ at $\displaystyle (4,9)$.

Therefore, the minimum value of $\displaystyle f(4)$ is $\displaystyle 9.$

- Jul 27th 2006, 08:15 PMCaptainBlackQuote:

Originally Posted by**topsquark**

explanation the answer is obvious.

Since symbolic manipulation is error prone, what the answer emerges at

the end of a long chain of symbolic games leaves considerable doubt as

to the correctness of the argument without considerable checking.

Not that the rigorous derivation of the result is not needed some times

to confirm the obvious.

I have an amusing anecdote about this sort of thing, but I won't repeat it

here as I suspect I'm the only one who will find it amusing.

RonL