Question

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

• A. a maximum at $x=0$
• B. a minimum at $x=0$
• C. neither of two at $x=0$
• D. $f\left(x\right)$ is not derivable at $x=0$

Answer 1

Answer: (C)

neither of two at $x=0$

Let $f$ be a sufficiently differentiable function.
$f^{\prime }\left(c\right)=f^{\prime \prime }\left(c\right)=...=f^n\left(c\right)=0$ where $c$ is a number $n$ the domain of $f$ and $n$ is an integer, then one of the following is true:
$n$ is odd and we have a local extremum at c. More precisely:
1. $f^{n+1}\left(c\right)<0$, then $c$ is a point of a maximum
2. $f^{n+1}\left(c\right)<0$, then $c$ is a point of a minimum
Or
$n$ is even and we have a (local) saddle point at c. More precisely:
1. $f^{n+1}\left(c\right)<0$, then $c$ is a strictly decreasing point of inflection.
2. $f^{n+1}\left(c\right)<0$, then $c$ is a strictly increasing point of inflection.
We have $f^{\prime }\left(x\right)=(3-x).2e^{2x}-e^{2x}-4e^x-4x{e}^x-1=0$
For maxima or minima,
$f^{\prime }\left(x\right)=(5-2x)e^{2x}-4(1+x)e^x-1=0$
This is satisfied for $x=0$
Now, $f^{\prime \prime }\left(x\right)=(5-2x)\cdot 2e^{2x}-2e^{2x}-4e^x-4(1+x)e^x=10-2-4-4=0$ at $x=0$
So, we find $f^{\prime \prime \prime }\left(x\right),$ We have $f^{\prime \prime \prime }\left(x\right)=(8-4x)\cdot 2e^{2x}-4e^{2x}-4e^x-4(2+x)e^x=16-4-4-8=0$ at $x=0$
So we find the next differential coefficient $f^{iv}\left(x\right)$
We have $f^{iv}\left(x\right)=(12-8x)\cdot 2e^{2x}-8e^{2x}-4e^x-4(3+x)e^x=24-8-4-12=0$ at $x=0$
Now $f^v\left(x\right)=(12-8x)4e^{2x}-8\cdot 2e^{2x}-8.2e^{2x}-4e^x-4(3+x)e^x=48-16-16-4-4-12=-4\ne 0$ at $x=0$
Hence $f\left(x\right)$ has neither maximum nor minimum at $x=0.$
Ans: C