A police unit has deployed a tracking system on a highway with a speed of 65 mph. A driver passes through one radar detector at 2 pm and is traveling 60 mph at that moment. Then, the driver passes through a second radar detector 60 miles away at 2:45 pm, again traveling 60 mph at that moment. However, a speeding ticket is being issued for this driver. When he asked for an explanation the response was "Mean Value Theorem." explain. Your report should include :
1) brief explanation about the mean value Theorem.
2) detailed calculation steps.

The driver covered 60 miles in 50 minutes, so his average speed was more than 65 miles per hour (It was actually 72 mph). Thus, at some point during this interval, he must have been going faster than 65 mph

To prove this, assume he was going the maximum allowed speed of 65 mph throughout.

In that case he would not have been able to cover the 60 miles between the two radar points in 50 minutes.

Thus, at some point(s) during this interval he MUST have crossed the speed limit.

The mean value theorem essentially says that if we know the average rate of change over an interval, we know that the instantaneous rate of change must have crossed that average at some point in the interval, which seems intuitive to me.

answer: ♡♡♡♡♡♡♡♡♡♡

step-by-step explanation:

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.