First you learned (back in grammar school) that you can add, subtract, multiply, and divide numbers. Then you learned that you can add, subtract, multiply, and divide polynomials. Now you will learn that you can also add, subtract, multiply, and divide functions. Performing these operations on functions is no more complicated than the notation itself. For instance, when they give you the formulas for two functions and tell you to find the sum, all they're telling you to do is add the two formulas. There's nothing more to this topic than that, other than perhaps some simplification of the expressions involved.
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To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.
(f + g)(x) = f (x) + g(x)
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= [3x + 2] + [4 – 5x]
= 3x + 2 + 4 – 5x
= 3x – 5x + 2 + 4
= –2x + 6
(f – g)(x) = f (x) – g(x)
= [3x + 2] – [4 – 5x]
= 3x + 2 – 4 + 5x
= 3x + 5x + 2 – 4
= 8x – 2
(f × g)(x) = [f (x)][g(x)]
= (3x + 2)(4 – 5x)
= 12x + 8 – 15x2 – 10x
= –15x2 + 2x + 8
My answer is the neat listing of each of my results, clearly labelled as to which is which.
( f + g ) (x) = –2x + 6
( f – g ) (x) = 8x – 2
( f × g ) (x) = –15x2 + 2x + 8
(f /g)(x) = (3x + 2)/(4 – 5x)
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This exercise differs from the previous one in that I not only have to do the operations with the functions, but I also have to evaluate at a particular x-value. To find the answers, I can either work symbolically (like in the previous example) and then evaluate, or else I can find the values of the functions at x = 2 and then work from there. It's probably simpler in this case to evaluate first, so:
f (2) = 2(2) = 4
g(2) = (2) + 4 = 6
h(2) = 5 – (2)3 = 5 – 8 = –3
Now I can evaluate the listed expressions:
(f + g)(2) = f (2) + g(2)
(h – g)(2) = h(2) – g(2)
= –3 – 6 = –9
(f × h)(2) = f (2) × h(2)
(h / g)(2) = h(2) ÷ g(2)
= –3 ÷ 6 = –0.5
Then my answer is:
(f + g)(2) = 10, (h – g)(2) = –9, (f × h)(2) = –12, (h / g)(2) = –0.5
If you work symbolically first, and plug in the x-value only at the end, you'll still get the same results. Either way will work. Evaluating first is usually easier, but the choice is up to you.
You can use the Mathway widget below to practice operations on functions. Try the entered exercise, or type in your own exercise. Then click the button and select 'Solve' to compare your answer to Mathway's. (Or skip the widget and continue with the lesson.)
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(Clicking on 'Tap to view steps' on the widget's answer screen will take you to the Mathway site for a paid upgrade.)
This isn't really a functions-operations question, but something like this often arises in the functions-operations context. This looks much worse than it is, as long as I'm willing to take the time and be careful.
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The simplest way for me to proceed with this exercise is to work in pieces, simplifying as I go; then I'll put everything together and simplify at the end.
For the first part of the numerator, I need to plug the expression 'x + h' in for every 'x' in the formula for the function, using what I've learned about function notation, and then simplify:
f(x + h)
= 3(x + h)2 – (x + h) + 4
= 3(x2 + 2xh + h2) – x – h + 4
= 3x2 + 6xh + 3h2 – x – h + 4
The expression for the second part of the numerator is just the function itself:
Now I'll subtract and simplify:
f(x + h) – f(x)
= [3x2 + 6xh + 3h2 – x – h + 4] – [3x2 – x + 4]
= 3x2 + 6xh + 3h2 – x – h + 4 – 3x2 + x – 4
= 3x2 – 3x2 + 6xh + 3h2 – x + x – h + 4 – 4
= 6xh + 3h2 – h
All that remains is to divide by the denominator; factoring lets me simplify:
Now I'm supposed to evaluate at h = 0, so:
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6x + 3(0) – 1 = 6x – 1
simplified form: 6x + 3h – 1
value at h = 0: 6x – 1
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That's pretty much all there is to 'operations on functions' until you get to function composition. Don't let the notation for this topic worry you; it means nothing more than exactly what it says: add, subtract, multiply, or divide; then simplify and evaluate as necessary. Don't overthink this. It really is this simple.
Oh, and that last example? They put that in there so you can 'practice' stuff you'll be doing in calculus. You likely won't remember this by the time you actually get to calculus, but you'll follow a very similar process for finding something called 'derivatives'.
URL: https://www.purplemath.com/modules/fcnops.htm
Finding Domain and Range
Learning Objective(s)
·Find the domain of a square root function.
·Find the domain and range of a function from the algebraic form.
Functions are a correspondence between two sets, called the domain and the range. When defining a function, you usually state what kind of numbers the domain (x) and range (f(x)) values can be. But even if you say they are real numbers, that doesn’t mean that all real numbers can be used for x. It also doesn’t mean that all real numbers can be function values, f(x). There may be restrictions on the domain and range. The restrictions partly depend on the type of function.
In this topic, all functions will be restricted to real number values. That is, only real numbers can be used in the domain, and only real numbers can be in the range.
Restricting the domain
There are two main reasons why domains are restricted.
·You can’t divide by 0.
·You can’t take the square (or other even) root of a negative number, as the result will not be a real number.
In what kind of functions would these two issues occur?
Division by 0 could happen whenever the function has a variable in the denominator of a rational expression. That is, it’s something to look for in rational functions. Look at these examples, and note that “division by 0” doesn’t necessarily mean that x is 0!
Square roots of negative numbers could happen whenever the function has a variable under a radical with an even root. Look at these examples, and note that “square root of a negative variable” doesn’t necessarily mean that the value under the radical sign is negative! For example, if x = −4, then −x = −(−4) = 4, a positive number.
Remember, here the range is restricted to all real numbers. The range is also determined by the function and the domain. Consider these graphs, and think about what values of y are possible, and what values (if any) are not. In each case, the functions are real-valued—that is, x and f(x) can only be real numbers.
Remember the basic quadratic function: f(x) = x2 must always be positive, so f(x) ≥ 0 in that case. In general, quadratic functions always have a point with a maximum or greatest value (if it opens down) or a minimum or least value (it if opens up, like the one above). That means the range of a quadratic function will always be restricted to being above the minimum value or below the maximum value. For the function above, the range is f(x) ≥ −4.
Other polynomial functions with even degrees will have similar range restrictions. Polynomial functions with odd degrees, like f(x) = x3, will not have restrictions.
Square root functions look like half of a parabola, turned on its side. The fact that the square root portion must always be positive restricts the range of the basic function, , to only positive values. Changes to that function, such as the negative in front of the radical or the subtraction of 2, can change the range. The range of the function above is f(x) ≤ −2.
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Rational functions may seem tricky. There is nothing in the function that obviously restricts the range. However, rational functions have asymptotes—lines that the graph will get close to, but never cross or even touch. As you can see in the graph above, the domain restriction provides one asymptote, x = 6. The other is the line y = 1, which provides a restriction to the range. In this case, there are no values of x for which f(x) = 1. So, the range for this function is all real numbers except 1.
Finding domain and range of different functions is often a matter of asking yourself, what values can this function not have?
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You can check that the vertex is indeed at (1, 4). Since a quadratic function has two mirror image halves, the line of reflection has to be in the middle of two points with the same y value. The vertex must lie on the line of reflection, because it’s the only point that does not have a mirror image!
In the previous example, notice that when x = 2 and when x = 0, the function value is 1. (You can verify this by evaluating f(2) and f(0).) That is, both (2, 1) and (0, 1) are on the graph. The line of reflection here is x = 1, so the vertex must be at the point (1, f(1)). Evaluating f(1)gives f(1) = 4, so the vertex is at (1, 4).
You can check the horizontal asymptote, y = 3. Is it possible for to be equal to 3? Write an equation and try to solve it.
Since the attempt to solve ends with a false statement—0 cannot be equal to 6!—the equation has no solution. There is no value of x for which , so this proves that the range is restricted.
Summary
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Although a function may be given as “real valued,” it may be that the function has restrictions to its domain and range. There may be some real numbers that can’t be part of the domain or part of the range. This is particularly true with rational and radical functions, which can have restrictions to domain, range, or both. Other functions, such as quadratic functions and polynomial functions of even degree, also can have restrictions to their range.
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