Chapter 1, Problem 95CR

To determine

Tofind:The inverse of a function and verify given results involving inverse.

Answer to Problem 95CR

The inverse of function $f\left(x\right)=\frac{1}{2}x+3$ is $f^{-1}\left(x\right)=2\left(x-3\right)$ , and the results $f\left(f^{-1}\left(x\right)\right)=x$ and $f^{-1}\left(f\left(x\right)\right)=x$ stand verified.

Explanation of Solution

Given information:

The given function is function is $f\left(x\right)=\frac{1}{2}x+3$ . To verify results $f\left(f^{-1}\left(x\right)\right)=x$ and $f^{-1}\left(f\left(x\right)\right)=x$ .

Method used:

The inverse of function $y=f\left(x\right);f:A\to B$ is given by $f^{-1}:B\to A$ .

Calculations:

The given function is $f\left(x\right)=\frac{1}{2}x+3$ .

Let $y=f\left(x\right)\Rightarrow x=f^{-1}\left(y\right)$

Or

  $f^{-1}\left(y\right)=x$........(1)

Now, from given function

  $y=\frac{1}{2}x+3\Rightarrow x=2\left(y-3\right)$........(2)

From results (1) and (2)

  $\Rightarrow f^{-1}\left(y\right)=x=2\left(y-3\right)$

Or $f^{-1}\left(y\right)=2\left(y-3\right)$ , that is equivalent to $f^{-1}\left(x\right)=2\left(x-3\right)$ .

Thus, the inverse of function $f\left(x\right)=\frac{1}{2}x+3$ is given by $f^{-1}\left(x\right)=2\left(x-3\right)$ .

Verification of Results:

We have $f\left(x\right)=\frac{1}{2}x+3$ (given) and $f^{-1}\left(x\right)=2\left(x-3\right)$ (derived)

Now, $f\left(f^{-1}\left(x\right)\right)=f\left(2(x-3)\right)$$\because f^{-1}\left(x\right)=2\left(x-3\right)$

  $\Rightarrow f\left(f^{-1}\left(x\right)\right)=\frac{1}{2}\left(\underset{\_}{2(x-3)}\right)+3$$\because f\left(x\right)=\frac{1}{2}x+3$

  $\Rightarrow f\left(f^{-1}\left(x\right)\right)=\frac{1}{2}\times 2(x-3)+3$

  $\Rightarrow f\left(f^{-1}\left(x\right)\right)=x-3+3$

  $\Rightarrow f\left(f^{-1}\left(x\right)\right)=x$

Hence, the result $f\left(f^{-1}\left(x\right)\right)=x$ is verified.

Further,

  $f^{-1}\left(f\left(x\right)\right)=f^{-1}\left(\frac{1}{2}x+3\right)$$\because f\left(x\right)=\frac{1}{2}x+3$

  $\Rightarrow f^{-1}\left(f\left(x\right)\right)=f^{-1}\left(\frac{1}{2}x+3\right)=2\left\{\left(\frac{1}{2}x+3\right)-3\right\}$$\because f^{-1}\left(x\right)=2\left(x-3\right)$

  $\Rightarrow f^{-1}\left(f\left(x\right)\right)=2\left\{\frac{1}{2}x+3-3\right\}$

  $\Rightarrow f^{-1}\left(f\left(x\right)\right)=2\left\{\frac{1}{2}x\right\}=x$

Hence, the result $f^{-1}\left(f\left(x\right)\right)=x$ is verified.

Conclusion:

The inverse of function $f\left(x\right)=\frac{1}{2}x+3$ is $f^{-1}\left(x\right)=2\left(x-3\right)$ , and the results $f\left(f^{-1}\left(x\right)\right)=x$ and $f^{-1}\left(f\left(x\right)\right)=x$ stand verified.