CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$f(x)=\sqrt{x+3}$$ and $$g(x)={x}^{2}+1$$ be two real functions, then find $$f\circ g$$ and $$g\circ f$$.


Solution

$$f(x)=\sqrt{x+3}$$
For domain, $$x+3\ge 0$$
$$\Rightarrow$$  $$x\ge -3$$
Domain of $$f=[-3,\infty)$$
Range of $$f=(-3,\infty)$$
Similarly, range of $$g=(1,\infty)$$

Then, rangle of f is subset of domain $$g$$ and range of $$g$$ is subset of $$f$$.
$$\therefore$$  $$f\circ g$$ and $$g\circ f$$ exist.

$$\Rightarrow$$  $$f\circ g(x)=f[g(x)]$$
                     $$=f(x^2+1)$$  ......... [ Since, $$g(x)^2+1$$ ]
                     $$=\sqrt{x^2+1+3}$$ ......... [ Since, $$f(x)=\sqrt{x+3}$$ ]
                     $$=\sqrt{x^2+4}$$

$$\Rightarrow$$  $$g\circ f(x)=g[f(x)]$$
                      $$=g(\sqrt{x+3})$$.........  [ Since, $$f(x)=\sqrt{x+3}$$ ]
                      $$=(\sqrt{x+3})^2+1$$......... [ Since, $$g(x)^2+1$$ ]
                      $$=x+3+1$$
                      $$=x+4$$

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image