Question Number 39671 by KMA last updated on 09/Jul/18
$${Three}\:{function}\:{are}\:{given}\:{f}\left({x}\right)=\sqrt{{x}} \\ $$$${g}\left({x}\right)={x}+\mathrm{5}\:{and}\:{h}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{5}\:.{Express} \\ $$$${h}\left({x}\right)\:{interms}\:{of}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right). \\ $$
Answered by MrW3 last updated on 10/Jul/18
![h(x)=x^2 +5=x^2 −x+x+5=x(x−1)+(x+5) ⇒h(x)=f^2 (x)[f^2 (x)−1]+g(x) or h(x)=f^4 (x)+5](https://www.tinkutara.com/question/Q39673.png)
$${h}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{5}={x}^{\mathrm{2}} −{x}+{x}+\mathrm{5}={x}\left({x}−\mathrm{1}\right)+\left({x}+\mathrm{5}\right) \\ $$$$\Rightarrow{h}\left({x}\right)={f}^{\mathrm{2}} \left({x}\right)\left[{f}^{\mathrm{2}} \left({x}\right)−\mathrm{1}\right]+{g}\left({x}\right) \\ $$$${or} \\ $$$${h}\left({x}\right)={f}^{\mathrm{4}} \left({x}\right)+\mathrm{5} \\ $$