The area of a rectangle is . The length of the rectangle is . What is the width of the rectangle? What restrictions, if any, are on this value?
As you saw in the Review Queue, we have already dealt with adding, subtracting, and multiplying functions. To add and subtract, you combine like terms (see the Adding and Subtracting Polynomials concept). When multiplying, you either FOIL or use the “box” method (see the Multiplying Polynomials concept). When you add, subtract, or multiply functions, it is exactly the same as what you would do with polynomials, except for the notation. Notice, in the Review Queue, we didn’t write out the entire function, just , for example. Let’s continue:
Distribute the negative sign to the second function and combine like terms. Be careful! . Also, this new function, has a different domain and range that either or .
If and , find and . Determine any restrictions for .
Solution: First, even though the is not written along with the and , it can be implied that and represent and .
To divide the two functions, we will place over in a fraction.
To find the restriction(s) on this function, we need to determine what value(s) of make the denominator zero because we cannot divide by zero. In this case . Also, the domain of is only , because we cannot take the square root of a negative number. The portion of the domain where is not defined is also considered part of the restriction. Whenever there is a restriction on a function, list it next to the function, separated by a semi-colon. We will not write separately because it is included in .
Now we will introduce a new way to manipulate functions; composing them. When you compose two functions, we put one function into the other, where ever there is an . The notation can look like or , and is read “ of of ”. Let’s do an example.
Using and from Example A find and and any restrictions on the domains.
Solution: For , we are going to put into everywhere there is an -value.
Now, substitute in the actual function for .
To find the domain of , let’s determine where is defined. The radicand is equal to zero when or . Between 4 and -4, the function is not defined because the square root would be negative. Therefore, the domain is all real numbers; .
Now, to find , we would put into everywhere there is an -value.
Notice that . It is possible that and is a special case, addressed in the next concept. To find the domain of , we will determine where is defined. is a line, so we would think that the domain is all real numbers. However, while simplifying the composition, the square and square root canceled out. Therefore, any restriction on or would still exist. The domain would be all real numbers such that from the domain of . Whenever operations cancel, the original restrictions from the inner function still exist . As with the case of , no simplifying occurred, so the domain was unique to that function.
If and , find and the restrictions on the domain.
Solution: Recall that is another way of writing . Let’s plug into .
The final function, because is being raised to the power, which will always yield a positive answer. Therefore, even when is negative, the answer will be positive. For example, if , then . An absolute value function has no restrictions on the domain. This will always happen when even roots and powers cancel. The range of this function is going to be all positive real numbers because the absolute value is never negative.
Recall, the previous example, however. The restrictions, if there are any, from the inner function, , still exist. Because there are no restrictions on , the domain of remains all real numbers.
Intro Problem Revisit If we set g equal to and f equal to , to find the width, we need to find .
Therefore the width of the rectangle is , and is a restriction on the answer.
and . Find:
4. and the domain
1. is the product of and .
Both representations are correct. Discuss with your teacher how s/he would like you to leave your answer.
2. Subtract from and simplify, if possible.
3. Divide by . Don’t forget to include the restriction(s).
Recall the properties of exponents. Anytime there is a negative exponent, it should be moved into the denominator. We set each factor in the denominator equal to zero to find the restrictions.
4. is a composition function. Let’s plug into everywhere there is an .
The domain of is all real numbers except , because we cannot divide by zero. Therefore, the domain of is all real numbers except .
5. is a composite function on itself. We will plug into everywhere there is an .
- A value of the domain where cannot be defined.
- Composite Function
- A function, , such that , also written . When and are composed, we plug into everywhere there is an -value, resulting in a new function, . The domain of is the set of all -values that are in the domain of and .
For problems 1-8, use the following functions to form the indicated compositions and clearly indicate any restrictions to the domain of the composite function.
For problems 9-16, use the following functions to form the indicated compositions and clearly indicate any restrictions to the domain of the composite function.