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Given the function defined by the equation $f(x)=2 x^{2}-1$ determine (a) $f(0),$ (b) $f(-2),(\text { c) } f(1),(\text { d) } f(3 x), \text { (e) } f(x+h)$

(a) -1(b) 7(c) 1(d) $18 x^{2}-1$(e) $2 x^{2}+4 x h+2 h^{2}-1$

Algebra

Chapter 1

Functions and their Applications

Section 2

Basic Notions of Functions

Functions

Campbell University

Harvey Mudd College

Baylor 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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03:10

Given the function defined…

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07:26

Find the following for eac…

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for this problem. We have been given a function f of X equals two X squared minus one. And through this problem, we're going to evaluate this function at a few different values of X. So how do we go about doing that? Well, let's let's look at our first case f of zero. What f of zero means is I'm gonna go find my f function and everywhere where there's an X in that function, I'm going to substitute in the value in my parentheses, which in this case, is zero. So if I go up to my F function, I'm gonna plug in a zero for X. So that's gonna be two times zero squared minus one, which is just going to give me a value of negative one. What if I try a different number? Maybe f of negative, too. Well, again, I'm finding the F function, and I'm substituting negative to infer X. So in this case, that would be two times negative. Two squared minus one. Well, that's gonna be two times four, which is 88 minus one is seven. Okay, let's try a third number. What about F of one now that function we substitute one in for X. That's going to give us two minus one, which equals one. Now. We can also do this with numbers with things that aren't necessarily just numbers. What if I had F of three X? Well, just because I have a variable inside those parentheses, the process stays the same. I'm still going to go to the F function. And everywhere there's an X in the original function, I'm going to substitute in three X, So this is gonna be two times three X squared minus one. Well, three squared is nine times two. That's 18 X squared minus one. Now my final answer has a variable still in it. But that's okay, because that's what I put into my function. If I put in a number, I typically going to get out a number, as when I evaluated. If I'm plugging in a variable, I'm gonna end up with a variable, most likely on the when I get my results. So let's try one more. Let's do f of X plus h. When I plugged that in, that gives me too. Here's my substitution X plus H squared minus one, and let's get rid of some of these parentheses. Let's expand binomial expansion X Plus H Squared is gonna be X squared plus two x h plus H squared minus one. And let's get rid of our parentheses. Two. X squared plus four x h plus two h squared minus one. Again, I've got variables in my final answer, but that's okay, because that's what I put in. So this is for this function ffx. We have evaluated with numbers and with some expressions and found the results.

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