### Video Transcript

For the function π of π₯ equals π₯
minus one to the eighth power, find all the possible values of π that satisfy the
mean value theorem over the closed interval zero to two.

Remember, the mean value theorem
says that if π is a function which is continuous over the closed interval π to π
and differentiable at every point of the open interval π to π. Then thereβs a point π in that
open interval, such that π prime of π is equal to π of π minus π of π over π
minus π. Our π of π₯ is π₯ minus one to the
eighth power. And our closed interval runs from
zero to two. Weβre looking to find the value of
π such that the derivative of our function evaluated at π is equal to π of π
minus π of π over π minus π. Thatβs π of two minus π of zero
over two minus zero. So weβll do two things. Weβll evaluate this quotient. And weβll also find an expression
for the derivative of our function and evaluate it at π. π of two minus π of zero is two
minus one to the eighth power minus zero minus one to the eighth power. And we obtain that π of two minus
π of zero over two minus zero is zero divided by two, which is just zero.

Our next job is to find the
derivative of our function. And weβll use the general power
rule. This says that if π of π₯ is a
differentiable function and π is π constant real number, such that π of π₯ is
equal to π of π₯ to the power of π. Then the derivative of π of π₯, π
prime of π₯, is equal to π times π of π₯ to the power of π minus one times the
derivative of π of π₯. Thatβs π prime of π₯. The derivative of our function π₯
minus one to the eighth power is therefore eight times π₯ minus one to the seventh
power multiplied by the derivative of π₯ minus one, which is one. And so we see that π prime of π₯
is eight times π₯ minus one to the seventh power.

We see that π prime of π is eight
times π minus one to the seventh power. And remember, we found that π of
π minus π of π over π minus π is zero. So weβre going to set this equal to
zero and then solve for π. We divide by eight. And we see that π minus one to the
seventh power is zero. And we can take the seventh root of
both sides to see that π minus one is equal to zero, which means that π equals one
is the only value of π that satisfies the mean value theorem. Notice it falls in the closed
interval zero to two, as required.