Question

Find the minimum and maximum values of x5-5x4+5x3-10

Answer 1

$y = {x}^{5} - 5 {x}^{4} + 5 {x}^{3} - 10 \\ \frac{dy}{dx} = 5x {}^{4} - 20 {x}^{3} + 15 {x}^{2} = 0 \\ x {}^{2} (5 {x}^{2} - 20x + 15) = 0 \\ {x}^{2} ( {x}^{2} - 4x + 3) = 0 \\ x = 0 \\ x = 3 \\ x = 1 \\ check \: \\ \frac{ {d}^{2}y }{d {x}^{2} } at \: these \: points \:$

Answer 2

Answer and Explanation:

Given : Equation $x^5-5x^4+5x^3-10$

To find : The minimum and maximum values of the equation ?

Solution :

To find the maximum minimum points we find the first derivative and get the critical points then substitute that points in the second derivative.

Let $y=x^5-5x^4+5x^3-10$

Derivate w.r.t x,

$y'=5x^4-20x^3+15x^2$

For critical points put y'=0,

$5x^4-20x^3+15x^2=0$

$5x^2(x^2-4x+3)=0$

$5x^2(x^2-3x-x+3)=0$

$5x^2(x(x-3)-1(x-3))=0$

$5x^2(x-1)(x-3)=0$

$5x^2=0,x-1=0,x-3=0$

$x=0,1,3$

The second derivative is

$y''=20x^3-60x^2+30x$

Substitute the critical values,

At x=0,

$y''=20(0)^3-60(0)^2+30(0)$

$y''=0$

There is no maximum and no minimum value.

At x=1,

$y''=20(1)^3-60(1)^2+30(1)$  

$y''=20-60+30$

$y''=-10<0$

The value is maximum.

At x=0,

$y''=20(3)^3-60(3)^2+30(3)$

$y''=20(27)-60(9)+90$

$y''=540-540+90$

$y''=90>0$

The value is minimum.