# Find the least value of 4x(squared) + 3 only x is squared

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Let `y=4x^2+3`

Then `dy/dx=d/dx(4x^2+3)`

=`4d/dx(x^2)+d/dx3`

=`4.2x+0`

=`8x`

Now for maximum or minimum `dy/dx=0`

so, `8x=0`

or, `x=0`

Now `(d^2y)/(dx^2)=8`

The condition is that if `(d^2y)/(dx^2)>0 `

at given value of x it is a point of minimum.

Here we see that `(d^2y)/(dx^2)>0` at `x=0` .

So the minimum value of `y=4x^2+3` is `y=4.0+3`

i.e. `y=3` .

`y=4x^2+3 `

`` if you know calculus ,

`dy/dx=8x`

For minimum value dy/dx=0

x=0

`(d^2y)/(dx^2)}_{x=0}=8`

`>0`

x=0 gives minimum value.

minimum value=4.0+3

=3