# 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 $\displaystyle f$ is countinous on $\displaystyle [a,b]$ and diffrenciable on $\displaystyle (a,b)$ then there exists are number $\displaystyle a<c<b$ such as,
$\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}$
-----
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 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 $\displaystyle f$ is countinous on $\displaystyle [a,b]$ and diffrenciable on $\displaystyle (a,b)$ then there exists are number $\displaystyle a<c<b$ such as,
$\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}$
-----
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,
$\displaystyle \min{f(4)\}=3(2)+3=9$

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 $\displaystyle f(1) = 3$ 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 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