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Given the function defined by the equation $f(x)=3 x-5,$ determine(a) $f(0),$ (b) $f(-1),(\text { c) } f(2) \text { (d) } f(2 x), \text { (e) } f(x+h)$

(a) -5(b) -8(c) 1(d) $6 x-5$(e) $3 x+3 h-5$

Algebra

Chapter 1

Functions and their Applications

Section 2

Basic Notions of Functions

Functions

Missouri State University

Campbell University

Harvey Mudd College

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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01:27

for this problem. We've been given a function f of X equals three X minus five and we want to determine some values of dysfunction at some given exes. So I'm going to just put a little box around here so we don't lose it. I can keep referencing it. Let's do some evaluations. First. Let's do F of zero. What does that mean? Well, that means I'm gonna go to my function f and everywhere where I see an X. In that function, I'm going to substitute the value zero. So this becomes then three times zero minus five or just negative five. Let's try another one. How about f of negative one? Well, again, I go to my F function and I substitute negative one every time X appears so I'd have three times negative. One minus five. Well, negative three minus five gives me negative eight. Next. Let's try f of to same thing. We've substitute to infer X. So that gives me three times two minus five and six. Minus five is just one. Okay, one more. Here. How about f of two X? Well, it's just because I got a letter in there now the process is still the same. I still go to my F function. I find my variable X, and I'm gonna plug in what I have in my parentheses in this case two x So I'll have three times two x minus five. And to get rid of parentheses, it's six x minus five. It's OK that I've got an X in my final answer because I had an ex in my input, that is, I've just plugged into X every time that X appeared in my original function and for one more let's do F of X plus h. So again, back to my original function three times X plus H minus five. And to get rid of those parentheses, I would have three X plus three h minus five. So for my given function the's. This is the evaluation of that function at some specified values.

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