show that f (x) = |5x| has bo invetse function.
Answers
Answer:
Step-by-step explanation:
Before defining the inverse of a function we need to have the right mental image of function.
Consider the function f(x) = 2x + 1. We know how to evaluate f at 3, f(3) = 2*3 + 1 = 7. In this section it helps to think of f as transforming a 3 into a 7, and f transforms a 5 into an 11, etc.
Now that we think of f as "acting on" numbers and transforming them, we can define the inverse of f as the function that "undoes" what f did. In other words, the inverse of f needs to take 7 back to 3, and take -3 back to -2, etc.
Let g(x) = (x - 1)/2. Then g(7) = 3, g(-3) = -2, and g(11) = 5, so g seems to be undoing what f did, at least for these three values. To prove that g is the inverse of f we must show that this is true for any value of x in the domain of f. In other words, g must take f(x) back to x for all values of x in the domain of f. So, g(f(x)) = x must hold for all x in the domain of f. The way to check this condition is to see that the formula for g(f(x)) simplifies to x.
g(f(x)) = g(2x + 1) = (2x + 1 -1)/2 = 2x/2 = x.
This simplification shows that if we choose any number and let f act it, then applying g to the result recovers our original number. We also need to see that this process works in reverse, or that f also undoes what g does.
f(g(x)) = f((x - 1)/2) = 2(x - 1)/2 + 1 = x - 1 + 1 = x.
Letting f-1 denote the inverse of f, we have just shown that g = f-1.
Definition:
Let f and g be two functions. If
f(g(x)) = x and g(f(x)) = x,
then g is the inverse of f and f is the inverse of g.
Exercise 1:
(a) Open the Java Calculator and enter the formulas for f and g. Note that you take a cube root by raising to the (1/3), and you do need to enter the exponent as (1/3), and not a decimal approximation. So the text for the g box will be
(x - 2)^(1/3)
Use the calculator to evaluate f(g(4)) and g(f(-3)). g is the inverse of f, but due to round off error, the calculator may not return the exact value that you start with. Try f(g(-2)). The answers will vary for different computers. However, on our test machine f(g(4)) returned 4; g(f(-3)) returned 3; but, f(g(-2)) returned -1.9999999999999991, which is pretty close to -2.
The calculator can give us a good indication that g is the inverse of f, but we cannot check all possible values of x.
(b) Prove that g is the inverse of f by simplifying the formulas for f(g(x) and g(f(x)).
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Graphs of Inverse Functions
We have seen examples of reflections in the plane. The reflection of a point (a,b) about the x-axis is (a,-b), and the reflection of (a,b) about the y-axis is (-a,b). Now we want to reflect about the line y = x.
The reflection of the point (a,b) about the line y = x is the point (b,a).
Let f(x) = x3 + 2. Then f(2) = 10 and the point (2,10) is on the graph of f. The inverse of f must take 10 back to 2, i.e. f-1(10)=2, so the point (10,2) is on the graph of f-1. The point (10,2) is the reflection in the line y = x of the point (2,10). The same argument can be made for all points on the graphs of f and f-1.
The graph of f-1 is the reflection about the line y = x of the graph of f.
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Existence of an Inverse
Some functions do not have inverse functions. For example, consider f(x) = x2. There are two numbers that f takes to 4, f(2) = 4 and f(-2) = 4. If f had an inverse, then the fact that f(2) = 4 would imply that the inverse of f takes 4 back to 2. On the other hand, since f(-2) = 4, the inverse of f would have to take 4 to -2. Therefore, there is no function that is the inverse of f.
Look at the same problem in terms of graphs. If f had an inverse, then its graph would be the reflection of the graph of f about the line y = x. The graph of f and its reflection about y = x are drawn below.
Note that the reflected graph does not pass the vertical line test, so it is not the graph of a function.
This generalizes as follows: A function f has an inverse if and only if when its graph is reflected about the line y = x, the result is the graph of a function (passes the vertical line test). But this can be simplified. We can tell before we reflect the graph whether or not any vertical line will intersect more than once by looking at how horizontal lines intersect the original graph!
Horizontal Line Test
Let f be a function.