Question 248890
{{{f(x) = (5x-3)/2}}}
{{{g(x) = (2x+3)/(2x-1)}}}
A. Composition: g o f (x)
g o f (x) is short for g(f(x))
Compositions can be easy if you learn how to interpret function rules. The key is to look at the argument as just a placeholder. For example, given that q(x) = 3x -7, this tells us that the function q will take its argument, multiply it by 3, and then subtract 7. Function q will do this to <b>any</b> input it gets:
q(j) = 3j - 7
q(3) = 3(3) - 7
q(x+5) = 3(x+5) - 7
q(h(x)) = 3(h(x)) - 7
No matter what input we give function q, it will multiply it by 3 and then subtract 7. This is what q(x) = 3x - 7 tells us.
So if {{{g(x) = (2x+3)/(2x-1)}}} then
{{{g(f(x)) = (2(f(x))+3)/(2(f(x))-1)}}}
and since {{{f(x) = (5x-3)/2}}} this becomes:
{{{g(f(x)) = (2((5x-3)/2)+3)/(2((5x-3)/2)-1)}}}
No we just simplify:
{{{g(f(x)) = ((5x-3)+3)/((5x-3)-1)}}}
{{{g(f(x)) = 5x/(5x-4)}}}<br>
B/C. Inverses. The inverse of a relation (or function) is the relation you get when you swap the x's and y's. And this pretty much describes the way we find inverses. Take your function's equation, using "y" instead of the function notation, and swap the x's and y's. You now have the equation for the inverse. Normally we solve this equation for y and then replace y with the notation for function inverses: f^-1(x)
{{{f(x) = (5x-3)/2}}}
Use "y" instead of f(x)
{{{y = (5x-3)/2}}}
Swap:
{{{x = (5y-3)/2}}}
This is the equation for the inverse. Now we solve this for y. Multiply both sides by 2:
{{{2x = 5y-3}}}
Add 3 to each side:
{{{2x+3 = 5y}}}
Divide both sides by 5:
{{{(2x+3)/5 = y}}}
Replace y with the notation for the inverse:
{{{(2x+3)/5}}} = f^-1(x)<br>
{{{g(x) = (2x+3)/(2x-1)}}}
Use "y" instead of f(x)
{{{y = (2x+3)/(2x-1)}}}
Swap:
{{{x = (2y+3)/(2y-1)}}}
This is the equation for the inverse. Now we solve this for y. Multiply both sides by (2y-1):
{{{(2y-1)x = (2y+3)}}}
{{{2xy - x = 2y+3}}}
Gather the y terms on one side by subtracting 2y from each side:
{{{2xy - x -2y = 3}}}
Move the non-y terms to the other side by adding x:
{{{2xy-2y = x+3}}}
Factor out y:
{{{y(2x-2) = x+3}}}
Divide by (2x-2):
{{{y = (x+3)/(2x-2)}}}
Replace y with the notation for the inverse:
g^-1(x) = {{{(x+3)/(2x-2)}}}<br>
D. Domains. When you are asked to determine a domain, you start by assuming that the domain is all Real numbers and then you look for x values, if any, that you must exclude. Reasons to exclude x values include those that:<ul><li>make a denominator zero</li><li>make the radicand of an even-numbered root negative (For example, for {{{y = sqrt(x+2)}}} we must exclude any x value that is less than -2 because they result in the square root of a negative number which we cannot allow.)</li><li>make an argument to a logarithm zero or negative</li><li>make arguments to other functions which are undefined (e.g. {{{tan(pi/2)}}}).</li></ul>
For g(x) we have no even-numbered roots or logarithms or other functions with undefined arguments to be worried about. But we do have a denominator which could be zero for the right (actually wrong) value of x. So we have to find what value of x makes 2x-1 become zero. In other words, we need to solve:
2x-1 = 0
2x = 1
x = 1/2
This is the only x value that makes the denominator zero. So this is the only x value that we must exclude from the domain. So the domain of g(x) is all Real numbers except 1/2.<br>
E. Range. Since y values are determined by the x values, finding a range means we first have to know what the domain is. From the previous problem, we know that the only number x cannot be is 1/2. Now we try to figure out what values g(x) can be when x can be any number except 1/2. Looking at
{{{g(x) = (2x+3)/(2x-1)}}}
it may be difficult to determine what the range is. In this case it can be helpful to transform the equation into one that is easier to analyze. With g(x) we can use long division:
<pre>
            1
      _______
2x-1 / 2x + 3
       2x - 1
       ======
            4
</pre>
So {{{g(x) = (2x+3)/(2x-1) = 1 + 4/(2x-1)}}}. Now that the x is in just one place we can more easily analyze g(x). Obviously a 1 is a 1 but what values can 4/(2x-1) be?<ul><li>For large positive values of x the denominator also becomes large and positive. 4 over a large positive denominator makes the fraction a small positive number.</li><li>For large negative values of x the denominator also becomes large and negative. 4 over a large negative denominator makes the fraction a small negative number.</li><li>For x values just a little larger than 1/2 (like 0.50000001), the denominator becomes a very small positive fraction. 4 over a very small positive fraction is a very large positive number</li><li>For x values just a little smaller than 1/2 (like 0.4999999), the denominator becomes a very small negative fraction. 4 over a very small negative fraction is a very large negative number</li><li>The only value 4/(2x-1) <b>cannot</b> be is zero. The only way a fraction can be zero is if the numerator is zero and the numerator of 4/(2x-1) will never be zero no matter what x might be.</li></ul>
(As you can see, finding a range can involve a good understanding of how fractions work.)<br>
So 4/(2x-1) can turn out to be any number except zero. And since g(x) = 1 + 4/(2x-1), g(x) can be any number except 1+0 or 1. The range is all Real numbers except 1.