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Find maximum and minimum value of y=4x^3+19x^2-14x+3

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

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

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

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

`:.`At `x=1/3` and `x=-7/2` 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=1/3)``=24(1/3)+38`

`=8+38`

`=46` (positive)

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

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

`=-84+38`

`=-46` (negative)

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

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

`"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`

`"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`






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