Chapter 1.3, Problem 39E

To determine

To find:the given function $f\circ g\circ h$ .

Answer to Problem 39E

  $\left(f\circ g\circ h\right)\left(x\right)=\sqrt{x^6+4x^3+1}$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=\sqrt{x-3},g\left(x\right)=x^2,h\left(x\right)=x^3+2$ .

Calculation:

As the given function is $f\left(x\right)=2x-1,g\left(x\right)=x^2,h\left(x\right)=1-x$ .

  $\begin{array}{l}\left(f\circ g\circ h\right)\left(x\right)=f\left(g\left(h\left(x\right)\right)\right)\\ =f\sqrt{x^2-3}\{\text{as }g\left(x\right)=x^2\}\\ =\sqrt{{\left(x^3+2\right)}^2-3}\{\text{as }h\left(x\right)=x^3+2\}\\ =\sqrt{\left(x^6+4x^3+4\right)-3}\\ =\sqrt{x^6+4x^3+1}\end{array}$

Hence,the range and domain are $\left(f\circ g\circ h\right)\left(x\right)=\sqrt{x^6+4x^3+1}$