Question

If y =x^4 -4x^3 +6x^2 -4x -3, show that y has a minimum at x =1

Answer 1

Step-by-step explanation:

Given,

$y = {x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 4x - 3$

We have to show that minimum is at x = 1.

Now,

Differentiating both the sides wrt x,

We get,

$= > \frac{dy}{dx} = 4 {x}^{3} - 12 {x}^{2} + 12x - 4$

Now,

To check minima or maxima,

$\frac{dy}{dx}=0$

Therefore,

We get,

$=>4{x}^{3}-12{x}^{2}+12x-4=0$

Now,

Put the Value of x = 1

Therefore,

We get,

$=>4\times1-12\times1+12\times1-4=0\\\\=>0=0$

Therefore,

x = 1 is a point for maxima or minima.

Now,

When x is slightly less than 1,

Then clearly,

$\frac{dy}{dx}= -ve$

And,

When x is slightly more than 1,

Then Clearly,

$\frac{dy}{dx}=+ve$

Therefore,

$\frac{dy}{dx}$ changes sign from -ve to +ve as x passes through 1.

Hence,

x = 1 is point of minima.