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

Expert Answers
rakesh05 eNotes educator| Certified Educator

Let    `y=4x^2+3`

   Then `dy/dx=d/dx(4x^2+3)`




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` .

mariloucortez eNotes educator| Certified Educator
Let rewrite the given this way: y=4x^2 + 3. From that, you can say that it is an equation of the parabola. y=4x^2 + 3 is a parabola that opens upward, so you can conclude that the vertex of the parabola is the minimum value. Let's transform the equation in standard form of a parabola that opens upward. Here is the pattern: y = a(x-h) + k, where (h,k) is the vertex and the minimum value is k. By inspection of the equation, k = +3, so the minimum value of 4x^2 + 3 is +3.
pramodpandey | Student

`y=4x^2+3 `

`` if you know calculus ,


For minimum value   dy/dx=0




x=0 gives minimum value.

minimum value=4.0+3