Mean value theorem

[MarkSA]

Hello, I have the following two problems: (No work yet unfortunately because I can't figure out where to begin on these..)

1.) Suppose that 3 <= f'(x) <= 5 for all values of x. Show that 18 <= f(8) - f(2) <= 30.

2.) At 2:00pm, a car's speedometer reads 30mi/hr. At 2:10pm, it reads 50mi/hr. Show that at some time between 2:00 and 2:10 the acceleration is exactly 120mi/h^2.

Well, it mentions acceleration, so i'm guessing it's dealing with the second derivative (f'') somehow... beyond that I have no idea where to begin. Can anyone provide some tips? Thanks.

 

[galactus]

1.) Suppose that 3 <= f'(x) <= 5 for all values of x. Show that 18 <= f(8) - f(2) <= 30.

Notice the multiplication by 6?.

\(\displaystyle \L\\18\leq{f(8)-f(2)}\leq{30}\)

Divide by 6:

\(\displaystyle \L\\3\leq\frac{f(8)-f(2)}{6}\leq{5}\)

But \(\displaystyle \L\\\frac{f(8)-f(6)}{8-2}=f'(c)\)

 

[MarkSA]

Tricky! I thought the x 6 looked suspicious, but I wasn't sure where to take it from there. Thanks.

Any ideas on where to start with the second problem?

 

[Deleted member 4993]

Hello, I have the following two problems: (No work yet unfortunately because I can't figure out where to begin on these..)

1.) Suppose that 3 <= f'(x) <= 5 for all values of x. Show that 18 <= f(8) - f(2) <= 30.

2.) At 2:00pm, a car's speedometer reads 30mi/hr. At 2:10pm, it reads 50mi/hr. Show that at some time between 2:00 and 2:10 the acceleration is exactly 120mi/h^2.

Same principle as 1.

Notice 10 min = 1/6 hour

and

(50-30)*6 = 120

Well, it mentions acceleration, so i'm guessing it's dealing with the second derivative (f'') somehow... beyond that I have no idea where to begin. Can anyone provide some tips? Thanks.