Question

At $x=0,f\left(x\right)=(3-x)e^{2x}-4x{e}^x-x$

• A. Has a minimum
• B. Has a maximum
• C. Has no extremum
• D. Is not defined

Answer 1

Answer: (C)

Has no extremum

We have $f^{\prime }\left(x\right)=(3-x).2e^{2x}-e^{2x}-4e^x-4x{e}^x-1=0$
for maxima or minima. or $f^{\prime }\left(x\right)=(5-2x)e^{2x}-4(1+x)e^x-1=0$ This is satisfied for $x=0$
Now $f^{\prime \prime }\left(x\right)=(5-2x).2e^{2x}-2e^{2x}-4e^x-4(1+x)e^x=10-2-4-4=0$ at $x=0$
So we find $f^{\prime \prime \prime }\left(x\right),$ We have $f^{\prime \prime \prime }\left(x\right)=(8-4x).2e^{2x}-4e^{2x}-4e^x-4(2+x)e^x=16-4-4-8=0$ at $x=0$
So we find the next differential coefficient $f^{iv}\left(x\right)$
We have $f^{iv}\left(x\right)=(12-8x).2e^{2x}-8e^{2x}-4e^x-4(3+x)e^x=24-8-4-12=0$ at $x=0$
Now $f^v\left(x\right)=(12-8x)4e^{2x}-8.2e^{2x}-8.2e^{2x}-4e^x-4(3+x)e^x=48-16-16-4-4-12=-4\ne 0$ at $x=0$
Hence $f\left(x\right)$ has neither maximum nor minimum at $x=0.$
Ans: C