Question

Let $f\left(x\right)={(x-3)}^5{(x-1)}^4$ then

• A. $x=7/9$ is a point of maxima
• B. $x=3$ is a point of minimum
• C. $x=-1$ is a point of maxima
• D. $f$ has no point of maximum or minimum

Answer 1

The correct option is C $x=-1$ is a point of maxima

$f^{\prime }\left(x\right)={(x-3)}^4{(x+1)}^3(9x-7)$
and $f^{\prime \prime }\left(x\right)=8{(x-3)}^3{(x+1)}^2(9x^2+14x+1)$
$f^{\prime \prime }\left(x\right)=24{(x-3)}^2(x+1)(21x^3-49x^2+7x+13)$
$f^{iv}\left(x\right)=24(x-3)(3x-1)(21x^3-49x^2+7x+13)+168{(x+3)}^2(x+1)(9x^2-14x+13)$
$f^v\left(x\right)=48(3x-5)(21x^3-49x^2+7x+13)+336{(x-3)}^2(x+1)(9x-7)$
$f^{\prime }\left(3\right)=f^{\prime }(-1)=f^{\prime }\left(7/9\right)=0$
At $x=-1$, $f^v\left(x\right)<0$ so $x=-1$ is a point of maxima
At $x=3$, $f^{\prime \prime }\left(x\right)=f^{\prime \prime \prime }\left(x\right)=f^{iv}\left(x\right)=0$ but $f^v\left(x\right)\ne 0$
So $f$ has neither maximum nor minimum at $x=3$.
At $x=7/9$, $f^{\prime \prime }\left(x\right)>0$ and therefore is a point of minima.
Ans: C