Question

Suppose that f(0)=3 and f′(x)≤7 for all values of x. Use the Mean Value Theorem to determine how large f(4) can possibly be.

Answer 1

MVT says that in \(a < x < b\), there is a point \(c\) where the derivative is equal to the average slope:
$$
f'(c) = \frac{f(b) - f(a)}{b-a} \, .
$$
So certainly, \(f'(c) <= 7\) since the average slope is less than or equal to the maximum slope:
$$
\frac{f(b) - f(a)}{b-a} \, \le 7\\
f'(c)(b-a)=f(b) - f(a)\le 7(b-a)\\
\implies f(4)-f(0) \le 7(4-0)\\
\implies f(4) \le 28+f(0)=28+3=31\\
\implies f(4) \le 31
$$

Answer 2

Ok thanks. How would you do a problem like this one: Suppose that 2≤f′(x)≤4 for all values of x. Use the Mean Value Theorem to find values for the inequality below.
___≤f(2)−f(−2)≤___

Answer 3

@ybarrap

Answer 4

4 and 2 are the upper and lower bounds on the possible slopes of \(f(x)\). Let b=2 and a=-2 in the formula for the MVT:
$$
f'(c) = \frac{f(b) - f(a)}{b-a} \, .\\
2\le f'(c) = \frac{f(2) - f(-2)}{2-(-2)} \le 4.\\
\implies 2(2+2)\le f'(c) = f(4) - f(2) \le 4(2+2)\\
\implies 8 \le f(4) - f(2) \le 16
$$

Answer 5

ok thanks. how would you do something like this: At 2:00pm a car's speedometer reads 20mph, and at 2:10pm it reads 40mph. Use the Mean Value Theorem to find an acceleration the car must achieve.

Answer 6

wouldn't f(2)-f(2)=0?

Answer 7

*few typos (see red), but everything else correct:
$$
f'(c) = \frac{f(b) - f(a)}{b-a} \, .\\
2\le f'(c) = \frac{f(2) - f(-2)}{2-(-2)} \le 4.\\
\implies 2(2+2)\le f'(c) = f(\color{red}2) - f(\color{red}{-2}) \le 4(2+2)\\
\implies 8 \le f(\color{red}2) - f(\color{red}{-2}) \le 16
$$

Answer 8

ok got it. now can you help me with At 2:00pm a car's speedometer reads 20mph, and at 2:10pm it reads 40mph. Use the Mean Value Theorem to find an acceleration the car must achieve. please

Answer 9

Well, the car can accelerate and decelerate in an infinite number of ways between points a and b to go from 20 mph to 40 mph in 10 mins, but there will always be an average acceleration, right? This average will be the same no matter how the car accelerates and decelerates in this time period. Convince your self of this. What would be the average acceleration?

According to the MVT, there is some point in the journey where the car will have exactly this value. Now, if the car were going this average value (of acceleration) throughout the journey, then it would still start at 20 and end in 40 mph. So, the acceleration that the car \(\bf must\) achieve is this average acceleration:
$$
\qquad \qquad \qquad \qquad \qquad f'(c) = \frac{f(10) - f(0)}{10-0}=? .
$$

Note that \(f'(c)\) here is acceleration and not velocity, because the ratio in this formula is a difference in velocities over time, which is the definition of acceleration.

Answer 10

so i have to solve for f'(c)

Answer 11

There is only one value that it can be.

Answer 12

f(10)/10?

Answer 13

$$
f'(c) = \frac{f(10) - f(0)}{10-0}= \frac{40 - 20}{10-0}=?
$$

Answer 14

2

Answer 15

I'll let you answer that one :)

Answer 16

but 40-20=20 and 10-0=10 and 20/10=2

Answer 17

but it can't be 2

Answer 18

@ybarrap

Answer 19

why not?

Answer 20

this is online hw. it tells you if you are right or wrong. it said i was wrong

Answer 21

what about the other questions, what did it say?

Answer 22

the other ones were right

Answer 23

I think I see the problem -- units. We divided by minutes, not hours. What is 10 minutes in units of hours?

Answer 24

what do you mean What is 10 minutes in units of hours?

Answer 25

yes, how many hours is 10 minutes?

Answer 26

1/6

Answer 27

yep

Answer 28

so finish the problem

Answer 29

what is f(1/6)?

Answer 30

no, divide by 1/6 hours instead of 10 minutes in the formula above

Answer 31

oh 20/(1/6)=120

Answer 32

yup thats it thatnks

Answer 33

alright -- hope you understood. The MVT has some surprising uses don't you think? I know I'm always surprised, I never know what I'm going to get. Just start with the formula and go from there. good luck and you're welcome.