Question

Find the points at which the function $f$ given by $f\left(x\right)=(x-2{)}^4(x+1{)}^3$ has
(i) local maxima
(ii) local minima
(iii) point of inflexion

• A. $i)x=-1,ii)x=\frac{2}{7},iii)2$
• B. $i)x=2,ii)x=\frac{2}{7},iii)-1$
• C. $i)x=\frac{2}{7},ii)x=2,iii)-1$
• D. $i)x=\frac{2}{7},ii)x=-1,iii)2$

Answer 1

The correct option is B $i)x=2,ii)x=\frac{2}{7},iii)-1$

Differentiate $f\left(x\right)={\left(x-2\right)}^4{\left(x+1\right)}^3$ with respect to $x$,

$f^{\prime }\left(x\right)=4{\left(x-2\right)}^3{\left(x+1\right)}^3+3{\left(x+1\right)}^2{\left(x-2\right)}^4$

$={\left(x-2\right)}^3{\left(x+1\right)}^2\left[4\left(x+1\right)+3\left(x-2\right)\right]$

$={\left(x-2\right)}^3{\left(x+1\right)}^2\left(4x+4+3x-6\right)$

$={\left(x-2\right)}^3{\left(x+1\right)}^2\left(7x-2\right)$

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

${\left(x-2\right)}^3{\left(x+1\right)}^2\left(7x-2\right)=0$

${\left(x-2\right)}^3=0$

$x=2$

Or,

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

$x=-1$

Or,

$\left(7x-2\right)=0$

$x=\frac{2}{7}$

At $x=2$,

When $x<2$, then, $f\left(x\right)<0$ and when $x>2$, then $f\left(x\right)>0$.

Since the sign changes from negative to positive, so the function has local maxima at $x=2$.

At $x=\frac{2}{7}$,

When $x<\frac{2}{7}$, then, $f\left(x\right)>0$ and when $x>\frac{2}{7}$, then $f\left(x\right)<0$.

Since the sign changes from positive to negative, so the function has local minima at $x=\frac{2}{7}$.

At $x=-1$,

When $x<-1$, then, $f\left(x\right)>0$ and when $x>-1$, then $f\left(x\right)>0$.

Since the sign does not changes, so the function has point of inflexion at $x=-1$.