Applications Of Derivatives Question 12
Question: What is the interval in which the function $ f(x)=\sqrt{9-x^{2}} $ is increasing- $ (f(x)>0) $
Options:
A) $ 0<x<3 $
B) $ -3<x<0 $
C) $ 0<x<9 $
D) $ -3<x<3 $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ f(x)=\sqrt{9-x^{2}} $
$ f’(x)=\frac{1}{2\sqrt{9-x^{2}}}\times (-2x)=-\frac{x}{\sqrt{9-x^{2}}} $ For function to be increasing $ -\frac{x}{\sqrt{9-x^{2}}}>0 $ or $ -x>0 $ or $ x<0 $ but $ \sqrt{9-x^{2}} $ is defined only when $ 9-x^{2}>0 $ or $ x^{2}-9<0 $
$ (x+3)(x-3)<0 $ i.e., $ -3<x<3 $
$ -3<x<3\cap x<0\Rightarrow -3<x<0 $
BETA
