Examples on Inverses of Functions

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Example - 37

Prove that if \(f\) is increasing, \({f^{ - 1}}\) is increasing and if \(f\) is decreasing, \({f^{ - 1}}\)  is decreasing.

Solution: If \(f\)  is increasing, then \({x_1} < {x_2}\)  implies that  \(f({x_1}) < f({x_2})\) or \({y_1} < {y_2}\)

Now, just reverse this argument.

For  \({y_1} < {y_2}\), we should have \({x_1} < {x_2}\)    (from above).

Hence \({y_1} < {y_2}\)   \( \Rightarrow \) \({x_1} < {x_2}\), which is equivalent to \({f^{ - 1}}({y_1}) < {f^{ - 1}}({y_2})\)

\(∴ {y_1} < {y_2}\)   \( \Rightarrow \) \({f^{ - 1}}({y_1}) < {f^{ - 1}}({y_2})\)

\( \Rightarrow \)  \({f^{ - 1}}\) is increasing.

The decreasing case can be proved similarly

Note: For students who are not convinced, read the following:

(i)  Visualise \(f\) in the form of a mapping. If \(f\) is increasing,

(ii)   Consider two statements P and Q, where P implies Q, i.e., P \( \Rightarrow \) Q.

Therefore, not (Q)     \( \Rightarrow \)     not (P)

For example

(Dark clouds are forming in the sky)     \( \Rightarrow \)     (It is about to rain)

(It is not about to rain)     \( \Rightarrow \)     (Dark clouds are not forming in the sky)

Lets apply this to our case. Since f is increasing.

\({x_1} < {x_2}\)   \( \Rightarrow \)    \({y_1} < {y_2}\)   \(\left( {{\rm{where}}\,\,y = f(x),\,\,\,x = {f^{ - 1}}(y)} \right)\)

Hence                    not \(({y_1} < {y_2})\)   \( \Rightarrow \)          not \(({x_1} < {x_2})\)

                   or      \({y_1} > {y_2}\)       \( \Rightarrow \)      \({x_1} > {x_2}\)

                   or    \({y_1} > {y_2}\)          \( \Rightarrow \)      \({f^{ - 1}}({y_1}) > {f^{ - 1}}({y_2})\)

Hence, \({f^{ - 1}}\)  is increasing.

 

Example – 38

If  \(f(x) = \sin x + \cos x\)  and  \(g(x) = {x^2} - 1,\)  then find the interval in which \(g\left( {f(x)} \right)\)  is invertible.

Solution:

\(\begin{align} g\left( {f(x)} \right) &= g(sinx + cosx) = {(sinx + cosx)^2} - 1\\   &= 1 + 2\sin x\,\cos x - 1 = sin2x  \end{align}\)

We have seen that \(\text{sin}x\) is invertible in \(\begin{align}\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right].\end{align}\) Therefore, \(\text{sin2}x\) will be invertible in \(\begin{align}\left[ { - \frac{\pi }{4},\frac{\pi }{4}}\right].\end{align}\)

In fact, any function is invertible in all such intervals where it is one-one and onto. \(\text{Sin2}x\) will be invertible in infinitely many such intervals. But \(\begin{align}\left[ { - \frac{\pi }{4},\frac{\pi }{4}} \right]\end{align}\) can be considered the basic invertible domain for \(\text{sin2}x\) since this interval is centered around the origin.

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