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# Problem on functions

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(a) Let f : (-inf,0) U (0,inf) -> R be defined by f(x) = x + 1/x. Show that f has no inverse function.
(b) Let g : (0,inf) -> R be defined by g(x) = 1/x. Show that g has an inverse function.

Dec 9, 2023

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(a) f has no inverse function

An inverse function exists if and only if a function is bijective (one-to-one and onto). To show that f does not have an inverse function, we can demonstrate that it is not bijective.

Not one-to-one: Consider the following pairs of distinct numbers:

x=1 and y=−1

x=1 and y=1

Both pairs map to the same output, f(1)=f(1)=0, violating the one-to-one property.

Not onto: The function f cannot take on the value −1 for any real number x. This is because:

If x is positive, then x + 1/x > 1, which is always greater than -1.

If x is negative, then x + 1/x < -1, which is always less than -1.

Since f cannot take on the value -1, it is not onto.

Therefore, f does not have an inverse function.

(b) g has an inverse function

The function g(x) = 1/x is one-to-one and onto on the interval (0, inf).

One-to-one: For any distinct x1 and x2 in the interval (0, inf), we have:

g(x1) = 1/x1 ≠ 1/x2 = g(x2)

Therefore, g maps distinct inputs to distinct outputs, fulfilling the one-to-one property.

Onto: For any real number y in the range (0, inf), we can find a corresponding x in the interval (0, inf) such that:

g(x) = 1/x = y

Solving for x, we get:

x = 1/y

Since a solution exists for any y in the range, g is onto.

Because g is both one-to-one and onto on the interval (0, inf), it has an inverse function. This inverse function is h(y) = 1/y, which is defined on the range (0, inf) and maps back to the domain (0, inf).

Dec 9, 2023