Question

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

• A. x$=-1$ is point of minima
• B. x$=-1$ is point of maxima
• C. x$=7/9$ is a point of minima
• D. x$=-1$ is neither a point of maxima and minima.

Answer 1

Answer: (C), (D)

The correct options are
C x$=7/9$ is a point of minima
D x$=-1$ is neither a point of maxima and minima.

Let $y=(x-3{)}^5(x+1{)}^4$
taking log both side,
$\log y=5\log (x-5)+4\log (x+1)$
Differentiating both sides w.r.t $x$
$\frac{1}{y}.\frac{dy}{dx}=\frac{5}{x-5}+\frac{4}{x+1}$
$\frac{dy}{dx}=(x-3{)}^4(x+1{)}^3\left[5\right(x+1)+4(x-3\left)\right]=(x-3{)}^4(x+1{)}^3(9x-7)=0$ for extremum.
$\Rightarrow x=-1,\frac{7}{9},3$
Now check the sign of $y^{\prime }$, for $x<-1,y^{\prime }>0$
for $-1<x<7/9,y^{\prime }<0$ and for $x>7/9,y^{\prime }>0$
Hence, $x=7/9$ is point of local minima.