Find maximum and minimum value of `y=4x^3+19x^2-14x+3`
Solution:
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(3x-1)(2x+7) = 0`
`=>(3x-1) = 0" or "(2x+7) = 0`
`=>3x = 1" or "2x = -7`
`=>x = 1/3" or "x = -7/2`
The solution is
`x = (1/3),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`
`=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`
`=4*(-7/2)^3+19*(-7/2)^2-14*(-7/2)+3`
`=-343/2+931/4+49+3`
`=453/4`
This material is intended as a summary. Use your textbook for detail explanation.
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