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Problem: maximum and minimum value of y=4x^3+19x^2-14x+3 [ Calculator, Method and examples ]

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Your problem `->` maximum and minimum value of y=4x^3+19x^2-14x+3


Here, `y=4x^3+19x^2-14x+3`

`:. (dy)/(dx)=``d/(dx)(4x^3+19x^2-14x+3)`

`=d/(dx)(4x^3)+d/(dx)(19x^2)-d/(dx)(14x)+d/(dx)(3)`

`=12x^2+38x-14+0`

`=12x^2+38x-14`

For stationary values, `(dy)/(dx)=0`

`=>12x^2+38x-14=0`

`=>2(6x^2+19x-7)=0`

`=>2(6x^2-2x+21x-7)=0`

`=>2(2x(3x-1)+7(3x-1))=0`

`=>2(2x+7)(3x-1)=0`

`=>2x+7=0" or "3x-1=0`

`=>2x=-7" or "3x=1`

`=>x=-7/2" or "x=1/3`

`:.`At `x=-7/2` and `x=1/3` we get stationary values.

Now, `(d^2y)/(dx^2)=``=12x^2+38x-14`

`d/(dx)(12x^2+38x-14)`

`=d/(dx)(12x^2)+d/(dx)(38x)-d/(dx)(14)`

`=24x+38-0`

`=24x+38`

`((d^2y)/(dx^2))_(x=-7/2)``=24*(-7/2)+38`

`=-84+38`

`=-46` (negative)

`:.` At `x=-7/2` the function is maximum

`((d^2y)/(dx^2))_(x=1/3)``=24*(1/3)+38`

`=8+38`

`=46` (positive)

`:.` At `x=1/3` the function is minimum

Now, `y=4x^3+19x^2-14x+3`

`"putting " x=-7/2`

`y_(max)``=4*(-7/2)^3+19*(-7/2)^2-14*(-7/2)+3`

`=-343/2+931/4+49+3`

`=453/4`

`"putting " x=1/3`

`y_(min)``=4*(1/3)^3+19*(1/3)^2-14*(1/3)+3`

`=4/27+19/9-14/3+3`

`=16/27`








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