f(x)=1/(x-1), x>1, a=2; Show that f(x) is one-to-one and find the domain and range of f-1

Question

f(x)=1/(x-1), x>1, a=2

1. Show that f is one-to-one
2. Calculate f^-1(x) and state the domain and range of f^-1

Answer 1

f(x) = 1/(x-1)

The easiest way to show that a function is one-to-one is to show that it is increasing or decreasing throughout its entire domain. We're not always lucky enough to have a function that is increasing or decreasing throughout its entire domain, but this time we're lucky. Our function f(x) = 1/(x-1) is decreasing throughout its entire domain. (The domain of our f is all the numbers greater than 1.)

One way to see that f is decreasing is to take its derivative, and then show that this derivative is always negative.

f(x) = 1/(x-1), but to make the differentiation easier, let's write this as f(x) = (x-1)^-1.

Therefore f'(x) = -1(x-1)^-2.

To show that -(x-1)^-2 is negative, let's write it as -1/((x-1)²).

Now we can see that throughout the domain of f (i.e., whenever x > 1), this derivative -1/((x-1)²) is negative because it's a fraction with a negative numerator -1 and a positive denominator (x-1)².

Therefore f is decreasing.

Therefore f is one-to-one.

To see that f is decreasing without the bother of taking a derivative, let a and b be any two numbers in the domain of f, with a < b.

We have to show that f(a) > f(b). That will mean that the function f goes "downhill".

Now f(a) = 1/(a-1) and f(b) = 1/(b-1). We have to show that 1/(a-1) > 1/(b-1).

(Recall that a < b, so a-1 < b-1.)

But it is true that 1/(a-1) > 1/(b-1) because when you have a fraction with a positive numerator and denominator, and you make the denominator bigger (b-1 is bigger than a-1), the value of the fraction gets smaller. (After all, 1/3 is smaller than 1/2, isn't it?)

So f(a) = 1/(a-1) > 1/(b-1) = f(b), so the function f goes downhill.

In other words, f is decreasing.

Therefore f is one-to one.

Therefore f has an inverse function f^-1.

(Let's get the formula for this inverse function f^-1.

To compute our original function f(x) = 1/(x-1), we took x, subtracted 1, and took the reciprocal of the result.

Therefore to compute the inverse function f^-1, we will have to take x, take its reciprocal, and add 1.

In other words, f^-1(y) = 1/y + 1

They told us that the domain of f is all the numbers greater than 1, in other words the set {x∈R: x > 1}.

(I'm using uppercase R to stand for the set of all real numbers.)

Therefore the range of the inverse function f^-1 is the same set of numbers {x∈R: x > 1}.

(Let's verify this. If x is any number greater than 1, let y = 1/(x-1). Then x = f^-1(y), so x is indeed in the range of f^-1.

The range of f is all the positive numbers, in other words the set {x∈R: x > 0}.

(Let's verify this. If y is any positive number, let x = 1/y + 1. Then y = f(x), so y is indeed in the range of f.)

Therefore the domain of the inverse function f^-1 is the same set of numbers {x∈R: x > 0}.

Answer 2

1. A function is one-to-one if no two different inputs are sent to the same output.

To show that f is one-to-one, we need to check that f(a) = f(b) implies that a = b. Indeed:

f(a) = f(b)

implies 1/(a-1) = 1/(b-1)

implies a-1 = b-1

implies a = b

The first implication is just the definition of f. The second comes from taking reciprocals, and the third from adding 1 to both sides.

2. We know that applying the original function f to f^-1(x) gives back x. So, we set f(f^-1(x)) = x and solve for f^-1(x):

f(f^-1(x)) = x

implies 1/(f^-1(x) - 1) = x

implies f^-1(x)- 1 = 1/x

implies f^-1(x) = 1 + 1/x

The first implication is just the definition of f. The second comes from taking reciprocals, and the third from adding 1 to both sides.

The domain of f^-1(x) is the same as the range of f, which is x > 0. The range of f is the domain of f, which is x > 1.

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NOTES: (a) Another way to see that f is one-to-one is to calculate the derivative and to observe that it is always negative. (b) The fact that f is one-to-one on its domain guarantees that the inverse function exists.