# Question Video: Finding the Expression of a Function given the Expression of Its Slope and a Point on the Function Using Indefinite Integration Mathematics

The slope at the point (𝑥, 𝑦) on the graph of a function is 6𝑒^(𝑥) + 2. What is 𝑓(𝑥), given that 𝑓(ln 5) = 1?

02:30

### Video Transcript

The slope at the point 𝑥, 𝑦 on the graph of a function is six 𝑒 to the power of 𝑥 plus two. What is 𝑓 of 𝑥, given that 𝑓 of the natural log of five equals one?

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

Well, 𝑓 prime of 𝑥 is the derivative of our function. But we also know that the opposite of differentiating is integrating. And so, if we integrate 𝑓 prime of 𝑥 with respect to 𝑥, we will find a general solution for 𝑓 of 𝑥. We’ll then use the fact that 𝑓 of the natural log of five is equal to one to find a particular solution. So, let’s integrate six 𝑒 to the power of 𝑥 plus two.

We begin by recalling that the integral of 𝑒 to the power of 𝑥 is 𝑒 to the power of 𝑥. Then, the integral of two is two 𝑥. 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 𝑓 of 𝑥 equals six 𝑒 to the power of 𝑥 plus two 𝑥 plus some constant 𝑐. Now, we said that 𝑓 of the natural log of five is equal to one. In other words, when 𝑥 is equal to the natural log of five, 𝑓 of 𝑥 is equal to one. So, we replace 𝑓 of 𝑥 with one. And then, each time we see an 𝑥 in our equation, we replace that with the natural log of five.

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

We then go back to the general solution for 𝑓 of 𝑥, and we replace 𝑐 with negative 29 minus two times the natural log of five. And so, we found 𝑓 of 𝑥. It’s six 𝑒 to the power of 𝑥 plus two 𝑥 minus 29 minus two times the natural log of five.