STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

$\therefore$ $\frac{{dy}}{{dx}} = 5{x^4} - 20{x^3} + 15{x^2}$$ = \,\,5{x^2}({x^2} - 4x + 3)$

$ = \,\,5{x^2}(x - 3)\,(x - 1)$

$\frac{{dy}}{{dx}} = 0$, gives $x = 0,\,1,\,3$

Now, $\frac{{{d^2}y}}{{d{x^2}}} = 20{x^3} - 60{x^2} + 30x$ = $10x(2{x^2} - 6x + 3)$

and $\frac{{{d^3}y}}{{d{x^3}}} = 10(6{x^2} - 12x + 3)$

For $x = 0$: $\frac{{dy}}{{dx}} = 0,\,\frac{{{d^2}y}}{{d{x^2}}} = 0,\,\frac{{{d^3}y}}{{d{x^3}}} \ne 0$

$\therefore$ Neither minimum nor maximum

For$x = 1,\,\frac{{{d^2}y}}{{d{x^2}}} = - 10 = {\rm{negative}}$.

$\therefore$ Maximum value ${y_{{\rm{max}}{\rm{.}}}} = - 9$

For $x = 3,\,\frac{{{d^2}y}}{{d{x^2}}} = 90 = {\rm{positive}}$

$\therefore$ Minimum value ${y_{{\rm{min}}{\rm{.}}}} = - 37$.