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

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

02:30

Video Transcript

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

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

Well, 𝑓 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 𝑓 prime of π‘₯ with respect to π‘₯, we will find a general solution for 𝑓 of π‘₯. 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 π‘₯ plus two.

We begin by recalling that the integral of 𝑒 to the power of π‘₯ is 𝑒 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 𝑓 of π‘₯ equals six 𝑒 to the power of π‘₯ plus two π‘₯ plus some constant 𝑐. Now, we said that 𝑓 of the natural log of five is equal to one. In other words, when π‘₯ is equal to the natural log of five, 𝑓 of π‘₯ is equal to one. So, we replace 𝑓 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 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 π‘₯, and we replace 𝑐 with negative 29 minus two times the natural log of five. And so, we found 𝑓 of π‘₯. It’s six 𝑒 to the power of π‘₯ plus two π‘₯ minus 29 minus two times the natural log of five.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.