Question

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

Answer 1

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 $x=-1,2,\frac{2}{7}$

Double derivative of this function is given by

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

At $x=-1$

$i.e$ $f^{{}^{\prime \prime }}(-1)$

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

$=0$

Let's do a first derivative test at $x=\frac{2}{7}$

Value close to the it, to the left $f^{{}^{\prime }}\left(x\right)>0$ and to the right $f^{{}^{\prime }}\left(x\right)<0$ Which shows that this is the point of Maxima.

At $x=2$

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

$=0$

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

Rather such a point is called as point of inflection.