`f(x)=3x^2+12x-15`
Find Local maxima and minima of a function using first derivative test

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`

`:.` `x=-2`

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Step-3: Use the critical points to determine intervals
There are total 1 critical points, So we have 2 intervals
`(-oo,-2),(-2,oo)`

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Step-4: Determine the Sign of `f^'(x)` in each interval

1. For first interval `(-oo,-2)`, we choose `x=-3`

`f^'(-3)``=6*(-3)+12`

`=-18+12`

`=-6`` < 0`

2. For second interval `(-2,oo)`, we choose `x=-1`

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

`=-6+12`

`=6`` > 0`

Interval	x-value	`f^'(x)`	Negative or Positive
`(-oo,-2)`	`-3`	`f^'(-3)=-6`` < 0`	Negative
`(-2,oo)`	`-1`	`f^'(-1)=6`` > 0`	Positive

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Step-5: Conclude the nature of each critical point
At `x=-2`, `f^'(x)` changes from positive to negative, indicating a local minimum

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Step-6: Determine the function values at the critical points
Calculate `f(x)` at the critical points to find the maximum and minimum values

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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