Video: Finding the Inverse of a Function Containing an Exponential Function

The function 𝑓(π‘₯) = 2𝑒^(π‘₯) + 3 has an inverse of the form 𝑔(π‘₯) = ln(π‘Žπ‘₯ + 𝑏). What are the values of π‘Ž and 𝑏?

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Video Transcript

The function 𝑓 π‘₯ equals two 𝑒 to the power of π‘₯ plus three has an inverse of the form 𝑔 π‘₯ equals ln of π‘Žπ‘₯ plus 𝑏. What are the values of π‘Ž and 𝑏?

Well, first of all, I’m just gonna write our function 𝑓 π‘₯ equals two 𝑒 to the power of π‘₯ plus three as 𝑦 equals two 𝑒 to the power of π‘₯ plus three. Now what we want to do is we actually want to find the inverse of this function.

Well, the first thing that we do is we actually exchange the variables. So as you can see, I’ve actually swapped π‘₯ and 𝑦. So we now have π‘₯ is equal to two 𝑒 to the power of 𝑦 plus three.

And now what we want to do if we wanna find the inverse is actually to rearrange to express in terms of 𝑦. So first of all, I’m actually gonna subtract three from each side. So I get π‘₯ minus three is equal to two 𝑒 to the power of 𝑦. And then next I would divide each side by two. So I’ve got π‘₯ minus three over two is equal to 𝑒 to the power of 𝑦. And then I’m gonna take the natural log of each side, which is gonna give us ln of π‘₯ minus three over two is equal to 𝑦.

Okay, so now we actually want to make sure that it’s in the form 𝑔 π‘₯ is equal to ln π‘Žπ‘₯ plus 𝑏. So therefore, we get 𝑦 is equal to ln and then a half π‘₯ minus three over two, with 𝑦 being 𝑔 π‘₯.

So we’ve actually now found the inverse of the function two 𝑒 to the power of π‘₯ plus three. So therefore, we can say that when we find the inverse of the function two 𝑒 to the power of π‘₯ plus three and we have it in the form 𝑔 π‘₯ is equal to ln of π‘Ž π‘₯ plus 𝑏, then π‘Ž is gonna be equal to a half and 𝑏 is gonna be equal to negative three over two.

Okay, so now we’ve got a solution. But what we want to do is actually recap what we’ve done to actually solve the problem. So first of all, we actually exchanged the variables, so in this case swapped π‘₯ and 𝑦 around. And the reason we did that is because the inverse of a function is actually a reflection about the line 𝑦 equals π‘₯.

So therefore, the 𝑦- and π‘₯-coordinates will switch. So that’s why we exchanged the variables. And then we rearranged to express in terms of 𝑦. Then once we did that, our 𝑦 was actually our inverse function. And then we used this to put it into the form 𝑔 π‘₯ is equal to ln π‘Ž π‘₯ plus 𝑏 and found π‘Ž and 𝑏.

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