### Video Transcript

The slope at the point π₯, π¦ on
the graph of a function is six π to the power of π₯ plus two. What is π of π₯, given that π of
the natural log of five equals one?

Weβve been given information about
the slope at a general point π₯, π¦. And, actually, if we recall, we can
find an expression for the slope of the graph of a function π of π₯ by finding its
derivative. This means given a function π of
π₯, its derivative π prime of π₯ at any point π₯, π¦ here will be defined by six π
to the power of π₯ plus two. Now, what do we know about the
relationship between π of π₯ and π prime of π₯?

Well, π prime of π₯ is the
derivative of our function. But we also know that the opposite
of differentiating is integrating. And so, if we integrate π prime of
π₯ with respect to π₯, we will find a general solution for π of π₯. Weβll then use the fact that π of
the natural log of five is equal to one to find a particular solution. So, letβs integrate six π to the
power of π₯ plus two.

We begin by recalling that the
integral of π to the power of π₯ is π to the power of π₯. Then, the integral of two is two
π₯. Now, of course, this is an
indefinite integral. So, we need that constant of
integration. And we can say that our function is
defined as π of π₯ equals six π to the power of π₯ plus two π₯ plus some constant
π. Now, we said that π of the natural
log of five is equal to one. In other words, when π₯ is equal to
the natural log of five, π of π₯ is equal to one. So, we replace π of π₯ with
one. And then, each time we see an π₯ in
our equation, we replace that with the natural log of five.

So, one is equal to six π to the
power of the natural log of five plus two times the natural log of five plus π. Of course, π to the power of the
natural log of five is just five. So, six π to the power of the
natural log of five is six times five, which is 30. We now see that we can solve for π
by subtracting 30 and two times the natural log of five from both sides of our
equation. And when we do, we find that π is
equal to negative 29 minus two times the natural log of five.

We then go back to the general
solution for π of π₯, and we replace π with negative 29 minus two times the
natural log of five. And so, we found π of π₯. Itβs six π to the power of π₯ plus
two π₯ minus 29 minus two times the natural log of five.