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Let $y=f(x)$ denote the upper half of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1,$ and $y=g(x)$ its lower half. Determine (a) $f(2),(\text { b) } g(2),(\text { c) } f(-1), \text { (d) } g(-1)$

(a) $\frac{3 \sqrt{21}}{5}$(b) $-\frac{3 \sqrt{21}}{5}$(c) $\frac{6 \sqrt{6}}{5}$(d) $-\frac{6 \sqrt{6}}{5}$

Algebra

Chapter 1

Functions and their Applications

Section 5

The Circle

Functions

Campbell University

Oregon State University

Baylor University

Idaho State University

Lectures

01:43

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x^2. The output of a function f corresponding to an input x is denoted by f(x).

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for this problem. We have been given the equation of an ellipse, and we want Thio. Um, look at some functions based on this ellipse, but we have a problem. I'm just going to sketch in the lips here, just just a random ellipse. Ellipses are not functions. They fail the vertical line test, and you can just see that if you imagine a line cutting down through this ellipse, it's going to touch the lips in more than one place. So I can't just turn the lips into a function. But here's what I can dio I can cut the ellipse in half. Imagine just putting a line straight through the center of it, right on that major axis that splits my lips into two pieces. I have the upper piece, which I'm doing in red, and I have the lower half, which I'm doing in blue. Each one of those haves is a function the red pieces, a function and the blue pieces of function. So what we're going to do is we're going to take this ellipse, and we're gonna break it into two pieces. First, I'm going to define a function f of X and F of X is going to be the upper half that corresponds with the red piece that I have there. And I'm also to define a function g of X. We're gonna let g of x b, the lower half of the Ellipse. So in either case, we're going to need to solve for why and that will give us our two functions f and G so back to our original function to solve for why? Let's get rid of these denominators. 25 times nine is 225. So I'm going to multiply both sides by 225. That gives me X squared times nine. I'll have why squared Times 25 equaling 225. And now we saw for why 20 Wife 25 y squared equals 225 minus nine x squared. I'm going to divide by 25 save a step. I'm just gonna do it that way and then take the square root how the square root is going to give me a plus or minus right there. That's the key to my two pieces, plus or minus 225 minus nine X squared over 25 positive values are going to give me my red function. Those were my That's my upper half. And if I do the negative ones, that's the lower half. That's gonna be my blue function. So let me just write these out, and I'll just I'll just keep color coding them the same way F. Of X is going to be the positive values square root of 225 minus nine X squared and the square to 25. I could put that as a denominator of five g of X. The lower half is going to be my negative values 2 25 minus nine x squared over five thes air identical functions, with the exception of that plus or minus in the beginning. So if I find a value for F of some number, the value for G for that same input, it's just going to be the negative of it. So it makes it easy to find matching pairs in my functions. So let's try to find a few. Let's start with X equal in two. So I want to find f of to Well, that's gonna be the square root of 225. Well, two squared is four. That's gonna B minus 36/5. Okay, 225 minus 36 is 189. And while that's not a perfect square, it does have a factor of nine. So I can pull that out to be a three. And I've got 21 left over five, so that gives me three times the square root of 21/5. Well, if I want to find G F two well, all the math is exactly the same. I just have to put a negative on the fronts. That's gonna be negative. Three times the square root of 21 over five. Okay, let's try one more value for X. Let's do X. Equally negative. One so f of negative one. That's the square root of 225. Well, one squared is one times nine. All that's gonna be over 5, 225 minus nine is 216 and 216 is actually six cube. So I have a six square toe underneath there. I could pull out of six. I'm left with one more factor of six under the radical so six times the square root of 6/5. And if I want to find the corresponding G again when X is negative, one is the same value, just with a negative. Okay, so those are my two functions that define together. They define this ellipse and some different values of the different functions at some specified values for X.

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