Question

At which of the following points does $f\left(x\right)=x^4$ has a maximum?

• A. 0
• B. 1
• C. 2
• D. 3
• E. None of these

Answer 1

The correct option is E

None of these

We saw that the necessary condition for $f\left(x\right)$ to have a maximum at $x=cis{f}^{\prime }\left(c\right)=0$

We have $f^{\prime }\left(x\right)=4x^3$

$f^{\prime }\left(x\right)=0\Rightarrow x=0$. Now we have to check if $x=0$ is a local maxima. For that, we have to check with nth derivative test conditions -

If $f\left(x\right)$ has derivative upto nth order and $f^{\prime }\left(c\right)=f”\left(c\right)..f^{n-1}\left(c\right)=0$, then

A) n is even, $f^n\left(c\right)<0\Rightarrow x=c$ is a point of maximum

B) n is even, $f^n\left(c\right)>0\Rightarrow x=c$ is a point of minimum

C) n is odd, $f^n\left(c\right)<0\Rightarrow f\left(x\right)$ is decreasing about $x=c$

D) n is odd, $f^n\left(c\right)>0\Rightarrow f\left(x\right)$ is increasing about $x=c$

So, we will differentiate the given function until we get a non negative value at $x=0$

$f\left(x\right)=x^4$

$f^{\prime }\left(x\right)=4x^3$

$f”\left(x\right)=12x^2$

$f”\left(0\right)=0$

$f{”}^{\prime }\left(x\right)=f^3\left(x\right)=24x$

$f^3\left(0\right)=0$

$f””\left(x\right)=f^4\left(x\right)=24$

$f^4\left(x\right)=24$

n is even, $f^n\left(c\right)>0\Rightarrow x=c$ is a point of minimum

$\Rightarrow x=0$ is a local minimum. So this function does not have any local maximum.