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

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seerboldly | Student, Grade 10 | (Level 1) Salutatorian

Posted March 30, 2013 at 12:05 PM via web

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

only x is squared 

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted March 30, 2013 at 12:25 PM (Answer #1)

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`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

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mariloucortez | High School Teacher | (Level 3) Adjunct Educator

Posted March 30, 2013 at 12:41 PM (Answer #2)

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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.
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rakesh05 | High School Teacher | (Level 1) Assistant Educator

Posted March 31, 2013 at 8:58 AM (Answer #3)

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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` .

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