Method
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Method of finding maximum or minimum values of a function.
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Step-1:
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Find the first derivative `(dy)/(dx)` of the function.
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Step-2:
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Put `(dy)/(dx)=0`, solve this equation and find the values of x.
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Step-3:
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These values of x give stationary points.
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Step-4:
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Find the secpmd derivative `(d^2y)/(dx^2)` of the function.
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Step-5:
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Put these values of x in the second derivative.
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Step-6:
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If `f''(x)<0` then it gives maximum value and If `f''(x)>0` then it gives minimum value.
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Example-1
Find maximum and minimum value of `y=x^3+6x^2-15x+7`
Solution:
Here, `y=x^3+6x^2-15x+7`
`:. (dy)/(dx)``d/(dx)(x^(3)+6x^(2)-15x+7)`
`=d/(dx)(x^(3))+d/(dx)(6x^(2))-d/(dx)(15x)+d/(dx)(7)`
`=3x^(2)+12x-15+0`
`=3x^(2)+12x-15`
For stationary values, `(dy)/(dx)=0`
`=>3x^(2)+12x-15=0`
`=>3(x^(2)+4x-5)=0`
`=>3(x^(2)-x+5x-5)=0`
`=>3(x(x-1)+5(x-1))=0`
`=>3(x+5)(x-1)=0`
`=>x+5=0" or "x-1=0`
`=>x=-5" or "x=1`
`:.`At `x=-5` and `x=1` we get stationary values.
Now, `(d^2y)/(dx^2)``d/(dx)(3x^(2)+12x-15)`
`=d/(dx)(3x^(2))+d/(dx)(12x)-d/(dx)(15)`
`=6x+12-0`
`=6x+12+0`
`=6x+12`
`((d^2y)/(dx^2))_(x=-5)``=6(-5)+12`
`=-30+12`
`=-18` (negative)
`:.` At `x=-5` the function is maximum
`((d^2y)/(dx^2))_(x=1)``=6*1+12`
`=6+12`
`=18` (positive)
`:.` At `x=1` the function is minimum
Now, `y=x^3+6x^2-15x+7`
`"putting " x=-5`
`y_(max)``=(-5)^(3)+6(-5)^(2)-15(-5)+7`
`=-125+150+75+7`
`=107`
`"putting " x=1`
`y_(min)``=1^(3)+6*1^(2)-15*1+7`
`=1+6-15+7`
`=-1`
This material is intended as a summary. Use your textbook for detail explanation.
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