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How many points uniquely determine (a) an exponential function?(b) a power function?

(a) 2(b) 2

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 2

Exponential Functions

Missouri State University

McMaster University

University of Michigan - Ann Arbor

Lectures

02:15

Write an exponential funct…

00:46

All exponential functions …

02:10

Find a formula for an expo…

03:56

03:10

01:05

Determine the exponential …

01:29

01:06

So I was looking through the book to see if in the chapter they just tell us how many points are needed to uniquely defined both these and I couldn't find anything. So what I decided to do is just kind of walk through how we can determine how many points are needed. So in each case, I'll just kind of tell you what the conclusion is going to be. We'll need two points or both. Um, but if you're not too content with that, we can actually prove that we need two points. So what I'm going to do is first say we have, um, two points, one that is going to be Let's call it R s and then another one that I'm going to call em in. So this first point here is saying that we have s is equal to a to the B R. And then the second point is saying we're going to have in is even to a be raised to the M like this now, one thing you might notice is if we divide both of these equations. So let me just divide them like this. The A's will cancel out and we'll end up with the following statements will be in over s is equal to be over and or be to them over B to the r and then be to them over B to the r. We would just attract the powers that be in over s was equal to He raised the M minus R. And now we can just raise each side to the one over in minus or power Or just take that route on each side and then notice we have b is equal to in over s raise to the one over. And my sir, So we have that. And now if we want to solve from A, we can just go ahead and plug this into either of those equations. So I'm just gonna plug it into the top one. Let me. Well, this Irish are just rewrite it. So we had in his equal to a B to the him Yeah, So first, let's just get a by itself solid Divide that over. So to be a is equal to in over B to the M and then be we found in the previous part to be, um, that big Montross city there. So be in over a not a and over in over s raised to the one over n minus r. And then all of this would be raised to the end power. And then remember, if we have two powers race to each other, we just multiply. So that is going to give us a is going to end over in over s raise to the M over in my star. And if you want, you could go ahead and reciprocate this and do all that. But I'm just going to be lazy and leave it like this. So what we found out is given any two points we can solve for A and B given these two equations where those original points are these here. So you can see how we used only two points to uniquely define something of the form. Why is it going to be raised to the X? Yeah, And now we're going to apply this same idea over here to show that we only need two points to uniquely show this. So, um, what points did I use again? I'll just use the same ones are s and M in. So we have our s. Mm. And? And we'll send it up just like we did before. So over here, this is going to say yes. Busy with two k times, our race to the people and then this is going to give us in is equal to k times him race the people. And now we can go ahead and divide each of these like we did before. So let me just go ahead and pick this stuff divided over here, and it doesn't matter which order we divide these, and this is just how I'm doing it. Um, the case cancel. So again we get in over us is going to be m to the P over our to the P. But now, this time we don't have the same base on the right side. So what we need to do is pull that power of P out. B n over s is equal to m over our raised to the P. And now, at this point,

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