`f(x)=3x^2+12x-15`
Find Local maxima and minima

Solution:
Here, `f(x)=3x^2+12x-15`

Step-1: Find the derivative of the function
`:. f^'(x)=``d/(dx)(3x^2+12x-15)`

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

`=6x+12-0`

`=6x+12`

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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`

`=>6x+12 = 0`

`=>6x = -12`

`=>x = (-12)/6`

`=>x = -2`

`:.` The critical points are `x=-2`

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Step-3: Apply the second derivative test
Now, `f^('')(x)=``d/(dx)(6x+12)`

`=d/(dx)(6x)+d/(dx)(12)`

`=6+0`

`=6`

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

For `x=-2`

`f^('')(-2)``=6`` > 0`

`:.` At `x=-2` the function is local minimum

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

`f(x)=3x^2+12x-15`

1. At `x=-2`

`f(-2)``=3*(-2)^2+12*(-2)-15`

`=12-24-15`

`=-27`

local minimum point = `(-2,-27)`

graph

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