Search any question & find its solution
Question:
Answered & Verified by Expert
If $f(x)=\sqrt{x}$ and $g(x)=x$ be two functions defined in the domain $R^{+} \cup\{0\}$, then find the value of.
(i) $(f+g)(x)$
(ii) $(f-g)(x)$
(iii) $(f g)(x)$
(iv) $\left(\frac{f}{g}\right)(x)$
(i) $(f+g)(x)$
(ii) $(f-g)(x)$
(iii) $(f g)(x)$
(iv) $\left(\frac{f}{g}\right)(x)$
Solution:
2048 Upvotes
Verified Answer
Given $f(x)=\sqrt{x}$ and $g(x)=x$
(i) $(f+g)(x)=f(x)+g(x)=\sqrt{x}+x$
(ii) $(f-g)(x)=f(x)-g(x)=\sqrt{x}-x$
(iii) $(f g)(x)=f(x) \cdot g(x)=\sqrt{x} \cdot x=x^{\frac{3}{2}}$
(iv) $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$
(i) $(f+g)(x)=f(x)+g(x)=\sqrt{x}+x$
(ii) $(f-g)(x)=f(x)-g(x)=\sqrt{x}-x$
(iii) $(f g)(x)=f(x) \cdot g(x)=\sqrt{x} \cdot x=x^{\frac{3}{2}}$
(iv) $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.