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

`:.` `x=-5` and `x=1`

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Step-3: Use the critical points to determine intervals
There are total 2 critical points, So we have 3 intervals
`(-oo,-5),(-5,1),(1,oo)`

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

1. For first interval `(-oo,-5)`, we choose `x=-6`

`f^'(-6)``=3*(-6)^2+12*(-6)-15`

`=108-72-15`

`=21`` > 0`

2. For second interval `(-5,1)`, we choose `x=0`

`f^'(0)``=3*0^2+12*0-15`

`=0+0-15`

`=-15`` < 0`

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

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

`=12+24-15`

`=21`` > 0`

Interval	x-value	`f^'(x)`	Negative or Positive
`(-oo,-5)`	`-6`	`f^'(-6)=21`` > 0`	Positive
`(-5,1)`	`0`	`f^'(0)=-15`` < 0`	Negative
`(1,oo)`	`2`	`f^'(2)=21`` > 0`	Positive

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

At `x=1`, `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=-5`

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

`=-125+150+75+7`

`=107`

Local Maximum point = `(-5,107)`

2. At `x=1`

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

`=1+6-15+7`

`=-1`

Local Minimum point = `(1,-1)`

graph

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