Question

If \(f\left( x \right) = {x^5} + {x^3} + x,\) find \({f^{ - 1}}\left( 3 \right)\) and \(f\left( {{f^{ - 1}}\left( 2 \right)} \right)\).

Short answer

\({f^{ - 1}}\left( 3 \right) = 1\) and \(f\left( {{f^{ - 1}}\left( 2 \right)} \right) = 2\).

Step-by-step solution

Condition for one-to-one function

If a function never takes on the same value again, it is called a one-to-one function. This means that;

\(f\left( {{x_1}} \right) \ne f\left( {{x_2}} \right)\) whenever \({x_1} \ne {x_2}\)

Determine \({f^{ - 1}}\left( 3 \right)\) and \(f\left( {{f^{ - 1}}\left( 2 \right)} \right)\)

To begin, we must find \(x\) such that \(f\left( x \right) = 3\).

According to the inspection, we can see that if \(x = 1\), then \(f\left( 1 \right) = 3\). So, \({f^{ - 1}}\left( 3 \right) = 1\) because \(f\) is a one-to-one function (and an increasing function), it has an inverse.

When \(f\) is a one-to-one function, then \(f\left( {{f^{ - 1}}\left( a \right)} \right) = a\), therefore, \(f\left( {{f^{ - 1}}\left( 2 \right)} \right) = 2\).

Thus, \({f^{ - 1}}\left( 3 \right) = 1\) and \(f\left( {{f^{ - 1}}\left( 2 \right)} \right) = 2\).