Remember that "x" is just a box, waiting for something to be put into it. Don't let odd-looking problems scare you. For instance:
Well, evaluating a function means plugging whatever they gave me in for the argument in the formula. This means that I have to plug this character "§" in for every instance of x. Here goes:
f (§) = (§)2 + 2(§) – 1
= §2 + 2§ – 1
This is definitely weird-looking, but it follows all the rules they gave me, so:
f (§) = §2 + 2§ – 1
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The above example is kinda pointless, I'll grant you, but it clearly illustrates how the notation works. You plug the given value in for the given variable, and chug your way to the answer. Hence, these exercises are often referred to as "plug-n-chug". Your best bet is to try not to over-think them.
This brings us to to our next topic: evaluating functions at variable expressions.
To evaluate this function at x = t, I'll need to plug t into every instance of x in the formula for the function g.
g(3) = 4 – (t)
= 4 – t
There's nothing more I can do with this, and I can't find a fully numerical value because I don't have a number to plug in for the t. So my answer is:
g(3) = 4 – t
Everywhere that my formula has an "x", I now plug in an "h". I start with the formula they gave me:
f (x) = 3x2 + 2x
If I want to be excruciatingly clear, I can start by writing the formula over again, this time with empty spaces where I'll be putting the new argument in place of the original variable:
f ( ) = 3( )2 + 2( )
Now I'll fill in those blanks with the new argument:
f (h) = 3(h)2 + 2(h)
There's nothing I can simplify, other than removing the extra parentheses. So my answer is:
f (h) = 3h2 + 2h
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Not every "evaluate at a variable expression" exercise is going to involve only variables. The variable expression can contain numbers, too.
For every instance of the variable w, I'll need to plug in the expression 2d + 1. I'll use parentheses to make the replacements clear for my next step.
h(2d + 1) = (2d + 1)2 – 3
I need to multiply out the squared binomial next, and then simplify:
(2d + 1)2 – 3
= (4d 2 + 4d + 1) – 3
= 4d 2 + 4d + 1 – 3
= 4d 2 + 4d – 2
I've simplified as much as I can. My answer is:
h(2d + 1) = 4d 2 + 4d – 2
Notice that, while doing the squaring halfway through my work above, I didn't ignore (and then maybe forget) the "minus three" that was along for the ride. In whatever manner you do your work, make sure that you're doing it so you don't get so busy with one part of an exercise that you find yourself in danger of losing track of other parts.
This one feels wrong, because it's asking me to plug something that involves x in for the original x. But this evaluation works exactly like all the others; namely, everywhere that the original formula has an "x", I will now plug in an "x + h".
f (x + h) = 3(x + h)2 + 2(x + h)
= 3(x2 + 2xh + h2) + 2x + 2h
= 3x2 + 6xh + 3h2 + 2x + 2h
If you're not sure how I got the stuff inside the parentheses (the set that the 3 was multiplied through), then you'll want to review how to simplify with parentheses and how to do polynomial multiplication.
I should not try to do this all at once. Instead, I'll break this into smaller, more manageable pieces.
(I also note that this exercise uses the same function as the previous exercise, and one of the substitutions is the same, too. So I'm gonna cheat a bit and copy that exercise's result for f (x + h).)
This function difference is the original function subtracted from the result of the previous exercise, so:
f (x + h) – f (x)
= [3x2 + 6xh + 3h2 + 2x + 2h] – [3x2 + 2x]
= 3x2 + 6xh + 3h2 + 2x + 2h – 3x2 – 2x
= 3x2 – 3x2 + 6xh + 3h2 + 2x – 2x+ 2h
= 6xh + 3h2 + 2h
In the result above, notice that f (x + h) – f (x) does not equal f (x + h – x) = f (h). You cannot "simplify" the different functions' arguments in this manner. Addition or subtraction of functions is not the same as addition or subtraction of the functions' arguments. Again, the parentheses in function notation do not indicate multiplication.
(This type of functional expression is called a "difference quotient", and is actually something you will see again in calculus. I guess the reason this sort of exercise crops up so commonly in algebra is that they're trying to "prep" you.
(But, to be fair, it's not like anybody remembers these by the time they get to calculus, so it's really a lot of work for no real purpose, in my opinion. However, this type of problem is quite popular, so you should expect to need to know how to do it, and should expect to see one on the next test.)
My best course of action is to break this up into pieces. Helpfully, the previous two exercises already had me do the first two pieces, setting me up with the final expression for f (x + h) – f(x). All I really need to do here is the final division. My work looks like this:
When working on complicated exercises like the last example above, use caution and start by splitting the exercises into the smaller, simpler steps demonstrated across the last three exercises above, so that you can complete these problems successfully. It's not that these are "hard" really, so much as being "very prone to the making of silly mistakes". Help yourself by taking it slow and doing one small chunk at a time.
You can use the Mathway widget below to practice evaluating functions at variable expressions. Try the entered exercise, or type in your own exercise. Then click the button and select "Evaluate" to compare your answer to Mathway's.
(Click on "Tap to view steps" to be taken directly to the Mathway site for a paid upgrade.)