Yulia

2021-01-10

Compose the following function and state the domain of the composed function.
$f\left(x\right)=\frac{1}{x}-1$
$g\left(x\right)=\sqrt{1-{x}^{2}}$
a) $g\left(f\left(-2\right)\right)$

Alannej

To find g(f(−2)), substitute -2 for x in the function $f\left(x\right)=\frac{1}{x-1}$ and then $-\frac{1}{3}$ for x in the function $g\left(x\right)=\sqrt{1-{x}^{2}}$, and simplify.
$g\left(f\left(-2\right)\right)=g\left(\frac{1}{-2-1}\right)$
$=\sqrt{1-{\left(-\frac{1}{3}\right)}^{2}}$
$=\sqrt{1-\frac{1}{9}}$
$=\sqrt{\frac{9-1}{9}}$
$=\frac{\sqrt{8}}{3}$
$=\frac{2\sqrt{2}}{3}$
Therefore, $g\left(f\left(-2\right)\right)=\frac{2\sqrt{2}}{3}$
To obtain the composite function $\left(g\circ f\right)\left(x\right)$, substitute $f\left(x\right)=\frac{1}{x-1}$ and $g\left(x\right)=\sqrt{1-{x}^{2}}$ in the definition of composite function $\left(g\circ f\right)\left(x\right)=g\left[f\left(x\right)\right]$, and simplify.
$\left(g\circ f\right)\left(x\right)=g\left[f\left(x\right)\right]$
$=g|\frac{1}{x-1}|$
$=\sqrt{1-{\left(\frac{1}{x-1}\right)}^{2}}$
$=\sqrt{\frac{{\left(x-1\right)}^{2}-1}{{\left(x-1\right)}^{2}}}$
$=\left(\frac{\sqrt{{x}^{2}-2x+1-1}}{x-1}$
$=\frac{\sqrt{x\left(x-2\right)}}{x-1}$
Therefore, $g\left[f\left(x\right)\right]=\frac{\sqrt{x\left(x-2\right)}}{x-1}$
To determine the domain of the composite function $\left(g\circ f\right)\left(x\right)$, use the definition of the composite function $g\left[f\left(x\right)\right]=\frac{\sqrt{x\left(x-2\right)}}{x-1}$ and identify the values of x, for which the denominator is not equal to zero and radicand is positive.
$x\le 0$
$x-2\ge 0$
$x\ge 2$
$x-1\ne 0$
$x\ne 1$
Hence, the domain of the composite function $g\left[f\left(x\right)\right]=\frac{\sqrt{x\left(x-2\right)}}{x-1}$ is