Question

$f\left(x\right)=6x^5$ has a maximum at x = 0.

• A. True
• B. False

Answer 1

The correct option is B

False

If f(x) 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, $f^n\left(c\right)<0=>x=c$ is a point of maximum.

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

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

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

So, we will differentiate until we get a non-zero value at $x=0$

$f\left(x\right)=6x^5$

$f^{\prime }\left(x\right)=30x^4$

$f^{\prime }\left(0\right)=0$

$f”\left(x\right)=120x^3$

$f”\left(0\right)=0$

$f{”}^{\prime }\left(x\right)=f^3\left(x\right)=360x^2$

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

$f””\left(x\right)=f^4\left(x\right)=720x$

$f^4\left(x\right)=0$

$f”{”}^{\prime }\left(x\right)=f^5\left(x\right)=720$

$f^5\left(0\right)=720$

which is non –zero.

So, we get n = 5, which is odd.

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

That is f(x) is increasing at $x=0.$ It does not have a maximum or minimum at $x=0$