Question

Find the critical points of the following functions and test them for their maxima and minima.$y=(x-3{)}^2(x-2{)}^2.$

Answer 1

$y=(x-3{)}^2(x-2{)}^2$

$\frac{dy}{dx}=(x-3{)}^2\frac{d}{dx}(x-2{)}^2+(x-2{)}^2\frac{d}{dx}(x-3{)}^2$

[Using product Rule]

$=(x-3{)}^{2}2(x-2)+(x-2{)}^2\cdot 2(x-3)$

$=2(x-2)(x-3)[x-3+x-2]$

$=2(x-2)(x-3)(2x-5)$

for maximum and minimum $\frac{dy}{dx}=0$

so $x=2,3,\frac{5}{2}$

Second derivative test

$\frac{dy}{dx}-2[x^2-5x+6](2x-5)$

$=2[2x^3-5x^2-10x^2+25x+12x-30]$

$=4x^3-10x^2-20x^2+50x+24x-30]$

at $x=2,x=3$ $\frac{d^{2}y}{dx}>0$ Point of local minima and

$\frac{d^{2}y}{dx}<0$ for $x=\frac{5}{2}$ Point of local maxima