# Calculus I Homework Help part II

• Jul 27th 2006, 12:21 PM
Nimmy
Calculus 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 PM
CaptainBlack
Quote:

Originally Posted by Nimmy
Given that f(1) = 3, and 2 <= f'(x) <= 3 for all x, what is the smallest possible value for f(4)?

f inceases most slowly if f'(x)=2, so the smallest possible value for f(4) is:

f(1)+3*2=9

RonL
• Jul 27th 2006, 12:57 PM
ThePerfectHacker
Quote:

Originally Posted by Nimmy
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)?

Let me give a more reasonable explanation.
Honestly, I do not understand what CaptainBlank is doing.

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

If a function $f$ is countinous on $[a,b]$ and diffrenciable on $(a,b)$ then there exists are number $a such as,
$f'(c)=\frac{f(b)-f(a)}{b-a}$
-----
Now comes the answer.
Since $2\leq f'(x)\leq 3$ for all $x$ you know that the function is diffrenciable for any $x$ (because it exists between 2 and 3). Furthermore, the function is countinous for any $x$ because diffrenciability implies counintuity.
Consider $f(x)$ on the interval $[1,4]$. By the previous explanation we have shown this function satisfies the conditions of the mean value theorem. Thus there is a $1 such as,
$f'(c)=\frac{f(4)-f(1)}{4-1}$. Thus,
Thus,
$3f'(c)=f(4)-f(1)$
Thus,
$3f'(c)=f(4)-3$
Thus,
$f(4)=3f'(c)+3$
Therefore the largest possible value of $f(4)$ is the largest possible value of $f'(c)$ which is 3. Thus,
$\max\{f(4)\}=3(3)+3=12$

And the smallest possible values is similarly,
$\min{f(4)\}=3(2)+3=9$
• Jul 27th 2006, 01:44 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Let me give a more reasonable explanation.
Honestly, I do not understand what CaptainBlank is doing.

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

If a function $f$ is countinous on $[a,b]$ and diffrenciable on $(a,b)$ then there exists are number $a such as,
$f'(c)=\frac{f(b)-f(a)}{b-a}$
-----
Now comes the answer.
Since $2\leq f'(x)\leq 3$ for all $x$ you know that the function is diffrenciable for any $x$ (because it exists between 2 and 3). Furthermore, the function is countinous for any $x$ because diffrenciability implies counintuity.
Consider $f(x)$ on the interval $[1,4]$. By the previous explanation we have shown this function satisfies the conditions of the mean value theorem. Thus there is a $1 such as,
$f'(c)=\frac{f(4)-f(1)}{4-1}$. Thus,
Thus,
$3f'(c)=f(4)-f(1)$
Thus,
$3f'(c)=f(4)-3$
Thus,
$f(4)=3f'(c)+3$
Therefore the largest possible value of $f(4)$ is the largest possible value of $f'(c)$ which is 3. Thus,
$\max\{f(4)\}=3(3)+3=12$

And the smallest possible values is similarly,
$\min{f(4)\}=3(2)+3=9$

Odd, I thought CaptainBlack's method was much simpler. I had to read yours twice to see what you were doing. ;)

-Dan
• Jul 27th 2006, 01:59 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Odd, I thought CaptainBlank's method was much simpler. I had to read yours twice to see what you were doing. ;)

-Dan

I guess my eye is trained to look only formally. (I still do not understand it :confused: ). Can you explain it to me. I am sure the Captain used a useful propery.
• Jul 27th 2006, 02:06 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
I guess my eye is trained to look only formally. (I still do not understand it :confused: ). Can you explain it to me. I am sure the Captain used a useful propery.

It is simply that the smallest possible value for f'(x) is 2 (over the interval [1,4] ) according to the problem statement, so the smallest possible change in f(x) from x = 1 to x = 4 is 2*(4-1) = 6.

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 PM
Soroban
Hello, Nimmy!

Quote:

Given that $f(1) = 3$ and $2 \leq f '(x) \leq 3$ for all $x$,
what is the smallest possible value for $f(4)$ ?

This is what the Captain meant . . .
Code:

      |              Q       |              * (4,?)       |              :       |              :       |  P          :       |  *          :       | (1,3)        :       |              :       |              :   - - + - + - + - + - + - -       |  1  2  3  4

We are given point $P(1,3)$
When $x = 4$, we have a point $Q(4,?)$

$2 \leq f'(x) \leq 3$ means:
. . the slope of the graph from $P$ to $Q$ is between $2$ and $3$.

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

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

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

• Jul 27th 2006, 08:15 PM
CaptainBlack
Quote:

Originally Posted by topsquark
It is simply that the smallest possible value for f'(x) is 2 (over the interval [1,4] ) according to the problem statement, so the smallest possible change in f(x) from x = 1 to x = 4 is 2*(4-1) = 6.

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

-Dan

Tsk, tsk! You aren't supposed to edit quotes! :p

I think my answer explains why the answer is what it is, with such an
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