Applications Of Derivatives Question 319
Question: The maximum value of $ x^{4}{e^{-x^{2}}} $ is
[AMU 1999]
Options:
A) $ e^{2} $
B) $ {e^{-2}} $
C) $ 12{e^{-2}} $
D) $ 4{e^{-2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x)=x^{4}{e^{-x^{2}}} $
therefore $ {f}’(x)=4x^{3}{e^{-x^{2}}}+x^{4}{e^{-x^{2}}}(-2x) $ For max., $ {f}’(x)=0 $
therefore $ 4x^{3}{e^{-x^{2}}}-2x^{5}{e^{-x^{2}}}=0 $
therefore $ x^{2}=2\Rightarrow x=\pm \sqrt{2} $
$ f’’(x)=12x^{2}{e^{-x^{2}}}+4x^{3}{e^{-x^{2}}}(-2x)-10x^{4}{e^{-x^{2}}}-2x^{5}{e^{-x^{2}}}(-2x) $
therefore $ {f}’’(\sqrt{2})=24{e^{-2}}-32{e^{-2}}-40{e^{-2}}+32{e^{-2}} $ = -ve
Hence, $ f(x) $ is max. at $ x=\sqrt{2} $
Maximum value = $ 4{e^{-2}} $ .
BETA
