Question

Find the values at which max. and min. value of the following exist.

$x^5-5x^4+5x^3-10.$

• A. $max=0,min=1$
• B. $max=0,min=3$
• C. $max=1,min=0$
• D. $max=1,min=3$

Answer 1

The correct option is D $max=1,min=3$

$y=x^5-5x^4+5x^3-10$
$\therefore \frac{dy}{dx}=5x^2(x^2-4x+3)$
$\frac{dy}{dx}=5x^2(x-1)(x-3)=0$ for min. or max
$\therefore x=0,1,3.$
Now $\frac{d^{2}y}{d{x}^2}=20x^3-60x^2+30x=10x(2x^2-6x+3)$
So at $x=0$, $\frac{d^{2}y}{d{x}^2}=0$, Hence point $x=0$ is neither maxima nor minima.
and ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=1}<0$ and ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=3}>0$
Thus point $x=1$ is maxima and $x=3$ is minima
$\therefore y_{max}=-9$ and $y_{min}=-37$