Operations on Functions
Question
Let \(f, g\) be two real functions defined by
\[ f(x)=\sqrt{x+1} \] \[ g(x)=\sqrt{9-x^2} \]Describe each of the following functions :
(i) \(f+g\)
(ii) \(g-f\)
(iii) \(fg\)
(iv) \(\frac{f}{g}\)
(v) \(\frac{g}{f}\)
(vi) \(2f-\sqrt{5}g\)
(vii) \(f^2+7f\)
(viii) \(\frac{5}{g}\)
Solution
Given
\[ f(x)=\sqrt{x+1} \] \[ g(x)=\sqrt{9-x^2} \]Domain of \(f(x)\):
\[ x+1\ge0 \] \[ x\ge -1 \]Therefore,
\[ D_f=[-1,\infty) \]Domain of \(g(x)\):
\[ 9-x^2\ge0 \] \[ x^2\le9 \] \[ -3\le x\le3 \]Therefore,
\[ D_g=[-3,3] \]Common domain:
\[ D_f \cap D_g=[-1,3] \](i) \(f+g\)
\[ (f+g)(x)=\sqrt{x+1}+\sqrt{9-x^2} \]Domain:
\[ [-1,3] \](ii) \(g-f\)
\[ (g-f)(x)=\sqrt{9-x^2}-\sqrt{x+1} \]Domain:
\[ [-1,3] \](iii) \(fg\)
\[ (fg)(x)=\sqrt{x+1}\sqrt{9-x^2} \]Domain:
\[ [-1,3] \](iv) \(\frac{f}{g}\)
\[ \left(\frac{f}{g}\right)(x)=\frac{\sqrt{x+1}}{\sqrt{9-x^2}} \]Since denominator cannot be zero,
\[ 9-x^2\ne0 \] \[ x\ne \pm3 \]From common domain \([-1,3]\), only \(x=3\) is excluded.
Domain:
\[ [-1,3) \](v) \(\frac{g}{f}\)
\[ \left(\frac{g}{f}\right)(x)=\frac{\sqrt{9-x^2}}{\sqrt{x+1}} \]Since denominator cannot be zero,
\[ x+1\ne0 \] \[ x\ne -1 \]Domain:
\[ (-1,3] \](vi) \(2f-\sqrt5g\)
\[ (2f-\sqrt5g)(x)=2\sqrt{x+1}-\sqrt5\sqrt{9-x^2} \]Domain:
\[ [-1,3] \](vii) \(f^2+7f\)
\[ (f^2+7f)(x)=(\sqrt{x+1})^2+7\sqrt{x+1} \] \[ =x+1+7\sqrt{x+1} \]Domain:
\[ [-1,\infty) \](viii) \(\frac{5}{g}\)
\[ \left(\frac{5}{g}\right)(x)=\frac{5}{\sqrt{9-x^2}} \]Denominator cannot be zero.
\[ 9-x^2\ne0 \] \[ x\ne \pm3 \]Domain:
\[ (-3,3) \]