Find maximum and minimum value of `y=3x^2+12x-15`
Solution:
Here, `y=3x^2+12x-15`
`:. (dy)/(dx)=``d/(dx)(3x^2+12x-15)`
`=d/(dx)(3x^2)+d/(dx)(12x)-d/(dx)(15)`
`=6x+12-0`
`=6x+12`
For stationary values, `(dy)/(dx)=0`
`=>6x+12=0`
`=>6x=0-12`
`=>6x=-12`
`=>x=(-12)/6`
`=>x=-2`
`:.`At `x=-2` we get stationary values.
Now, `(d^2y)/(dx^2)``=6x+12`
`d/(dx)(6x+12)`
`=d/(dx)(6x)+d/(dx)(12)`
`=6+0`
`=6`
`((d^2y)/(dx^2))_(x=-2)``=6` (positive)
`:.` At `x=-2` the function is minimum
Now, `y=3x^2+12x-15`
`"putting " x=-2`
`y_(min)``=3*(-2)^2+12*(-2)-15`
`=12-24-15`
`=-27`
This material is intended as a summary. Use your textbook for detail explanation.
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