Question

Find the point of maxima and minima of $f\left(x\right)=-{(x-1)}^3{(x+1)}^2$

• A. local maxima at $x=-\frac{1}{5},$ local minima at $x=-1$
• B. local maxima at $x=-\frac{1}{5}$, local minima at $x=-2$
• C. local maxima at $x=-\frac{1}{10}$, local minima at $x=-1$
• D. None of the above

Answer 1

The correct option is A local maxima at $x=-\frac{1}{5},$ local minima at $x=-1$

Given $f\left(x\right)=-(x-1{)}^3(x+1{)}^2$

finding the derivative and equating to zero to find the critical points.

$f^{{}^{\prime }}\left(x\right)=-3(x-1{)}^2(x+1{)}^2+(-(x-1{)}^3)\ast 2\ast (x+1)$

$⟹f^{{}^{\prime }}\left(x\right)=-3(x-1{)}^2(x+1{)}^2+(-2)(x-1{)}^3(x+1)$

$⟹f^{{}^{\prime }}\left(x\right)=(x-1{)}^2(x+1\left)\right(-3(x+1)+(-2)(x-1))$

$⟹f^{{}^{\prime }}\left(x\right)=(x-1{)}^2(x+1\left)\right(-3x-3-2x+2)$

$⟹f^{{}^{\prime }}\left(x\right)=(x-1{)}^2(x+1\left)\right(5x+1\left)\right(-1)$

$⟹(x-1{)}^2(x+1\left)\right(5x+1\left)\right(-1)=0$

$⟹(x-1{)}^2=0$ and $(x+1)=0$ and $(5x+1)=0$

$⟹x=1$ and $x=-1$ and $x=-\frac{1}{5}$

Therefore, $x=1,-1,-\frac{1}{5}$ are the critical points.

Now substituting the critical points in $f\left(x\right)$ to get the local maximum and the local minimum.

Therefore, $f\left(1\right)=-(1-1{)}^3(1+1{)}^2=0$

$f(-1)=-(-1-1{)}^3(-1+1{)}^2=0$

$f(-\frac{1}{5})=-(-\frac{1}{5}-1{)}^3(-\frac{1}{5}+1{)}^2=(-\frac{6}{5}{)}^3(-\frac{4}{5}{)}^2=-\frac{3456}{3125}=-1.1$

Hence, the local maximum occurs at $x=-\frac{1}{5}$ and local minimum occurs at $x=1orx=-1$