Chapter 1.2, Problem 32E

To determine

To find: $f\left(\frac{1}{2}\right)$ .

Answer to Problem 32E

  $\frac{1}{2}$

Explanation of Solution

Given information: $\begin{array}{c}g\left(x\right)=1-x^2\\ \left[f\circ g\right]\left[x\right]=\frac{x^4+x^2}{1+x^2}\end{array}$

Calculation:

Since, $\left[f\circ g\right]\left[x\right]=f\left(g\left(x\right)\right)$

Therefore, $\frac{x^4+x^2}{1+x^2}=f\left(g\left(x\right)\right)$

To find $f\left(\frac{1}{2}\right)$ , we have to guess the numerator and denominator $f\left(x\right)$ .

Let’s assume the denominator to be $\left(1+x\right)$ . Plug $g\left(x\right)$ in and we get,

  $1+\left(1-x^2\right)=2-x^2$

This value is not equal to the denominator of $\left[f\circ g\right]\left[x\right]$ .

Let’s assume the denominator to be $\left(1-x\right)$ . Plug $g\left(x\right)$ in and we get,

  $1-\left(1-x^2\right)=x^2$

This value is not equal to the denominator of $\left[f\circ g\right]\left[x\right]$ .

Let’s assume the denominator to be $\left(2-x\right)$ . Plug $g\left(x\right)$ in and we get,

  $2-\left(1-x^2\right)=1-x^2$

Since, this value is equal to the denominator of $\left[f\circ g\right]\left[x\right]$ . Thus, the denominator of $f\left(x\right)$ is $2-x$ .

Similarly, numerator of $f\left(x\right)$ can be determined.

Let’s assume the numerator to be $\left(1+x\right)$ . Plug $g\left(x\right)$ in and we get,

  $1+\left(1-x^2\right)=2-x^2$

This value is not equal to the numeratorof $\left[f\circ g\right]\left[x\right]$ .

Let’s assume the numeratorto be $x^2$ to get the power 4. Plug $g\left(x\right)$ in and we get,

  ${\left(1-x^2\right)}^4=1-2x^2+x^4$

This value is not equal to the numeratorof $\left[f\circ g\right]\left[x\right]$ .

Let’s assume the numeratorto be $\left(x^2+3x\right)$ to get 3x. Plug $g\left(x\right)$ in and we get,

  $\begin{array}{c}{\left(1-x^2\right)}^2+3\left(1-x^2\right)=1-2x^2+x^4+3-3x^2\\ =x^4-5x^2+4\end{array}$

This value is not equal to the numeratorof $\left[f\circ g\right]\left[x\right]$ .

Let’s assume the numeratorto be $\left(x^2-3x\right)$ to reverse the sign. Plug $g\left(x\right)$ in and we get,

  $\begin{array}{c}{\left(1-x^2\right)}^2-3\left(1-x^2\right)=1-2x^2+x^4-3+3x^2\\ =x^4+x^2-2\end{array}$

This value is not equal to the numeratorof $\left[f\circ g\right]\left[x\right]$ .

Let’s assume the numeratorto be $\left(x^2-3x+2\right)$ to remove +2. Plug $g\left(x\right)$ in and we get,

  $\begin{array}{c}{\left(1-x^2\right)}^2-3\left(1-x^2\right)+2=1-2x^2+x^4-3+3x^2+2\\ =x^4+x^2\end{array}$

Since, this value is equal to the numeratorof $\left[f\circ g\right]\left[x\right]$ . Thus, the numerator of $f\left(x\right)$ is $x^2-3x+2$ .

So, the function $f\left(x\right)$ can be written as follows:

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

Put $x=\frac{1}{2}$ in the above function, and evaluate $f\left(\frac{1}{2}\right)$ as shown below.

  $\begin{array}{c}f\left(\frac{1}{2}\right)=\frac{{\left(\frac{1}{2}\right)}^2-3\left(\frac{1}{2}\right)+2}{2-\left(\frac{1}{2}\right)}\\ =\frac{\left(\frac{1}{4}\right)-\left(\frac{3}{2}\right)+2}{2-\left(\frac{1}{2}\right)}\\ =\frac{\left(\frac{1}{4}\right)-\left(\frac{6}{4}\right)+\left(\frac{8}{4}\right)}{\left(\frac{4}{2}\right)-\left(\frac{1}{2}\right)}\\ =\frac{\left(\frac{3}{4}\right)}{\left(\frac{3}{2}\right)}\\ =\frac{1}{2}\end{array}$

Therefore, $f\left(\frac{1}{2}\right)$ is $\frac{1}{2}$ .