Question

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

Answer 1

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

Derivative of given function is

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

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

Putting this to zero we get $x=-1,2,\frac{2}{7}$

duble derivative of this function is given by

$f^{{}^{\prime \prime }}\left(x\right)=12{(x-2)}^2{(x+1)}^3+12{(x-2)}^3{(x+1)}^2+6(x+1){(x-2)}^4+12{(x+1)}^2{(x-2)}^3$

At $x=-1$ and $2$ double derivative is $0$ and hence function will not have any maxima or minima at these points.

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

$f^{{}^{\prime \prime }}\left(\frac{2}{7}\right)=12{(\frac{2}{7}-2)}^2{(\frac{2}{7}+1)}^3+12{(\frac{2}{7}-2)}^3{(\frac{2}{7}+1)}^2+6(\frac{2}{7}+1){(\frac{2}{7}-2)}^4+12{(\frac{2}{7}+1)}^2{(\frac{2}{7}-2)}^3$

$\Rightarrow f^{{}^{\prime \prime }}\left(\frac{2}{7}\right)<0$

So function will have local maxima at $x=\frac{2}{7}$