Method

Find Critical Points and Extrema (for single variable function)
Step-1:	Find the first derivative `f^'(x)` of the function.
Step-2:	Put `f^'(x)=0`, solve this equation and find the values of x.
Step-3:	These values of x give critical points.
Step-4:	Find the second derivative `f^('')(x)` of the function.
Step-5:	Put these values of x in the second derivative `f^('')(x)`.
Step-6:	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
`f(x)=x^3+6x^2-15x+7`
Find Second derivative test for Local maxima and minima

Solution:
Here, `f(x)=x^3+6x^2-15x+7`

Step-1: Find the derivative of the function
`:. f^'(x)=``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`

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Step-2: Find the critical points of the derivative function
To find critical points, set `f'(x)=0` and then solve for x

`f^'(x)=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-1)(x+5) = 0`

`=>(x-1) = 0" or "(x+5) = 0`

`=>x = 1" or "x = -5`

The solution is
`x = 1,x = -5`

`:.` The critical points are `x=1` and `x=-5`

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Step-3: Apply the second derivative test
Now, `f^('')(x)=``d/(dx)(3x^2+12x-15)`

`=d/(dx)(3x^2)+d/(dx)(12x)-d/(dx)(15)`

`=6x+12-0`

`=6x+12`

Evaluate `f^('')(x)` at the critical points

For `x=1`

`f^('')(1)``=6*1+12`

`=6+12`

`=18`` > 0`

`:.` At `x=1` the function is local minimum

For `x=-5`

`f^('')(-5)``=6*(-5)+12`

`=-30+12`

`=-18`` < 0`

`:.` At `x=-5` the function is local maximum

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Step-4: Calculate the extrema values
Substitute the `x` values back into the original function `f(x)`

`f(x)=x^3+6x^2-15x+7`

1. At `x=1`

`f(1)``=1^3+6*1^2-15*1+7`

`=1+6-15+7`

`=-1`

local minimum point = `(1,-1)`

2. At `x=-5`

`f(-5)``=(-5)^3+6*(-5)^2-15*(-5)+7`

`=-125+150+75+7`

`=107`

local maximum point = `(-5,107)`

graph

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