`f(x)=x^3-9x^2+24x+2`
Find Local maxima and minima of a function using first derivative test

Solution:
Here, `f(x)=x^3-9x^2+24x+2`

Step-1: Find the derivative of the function
`:. f^'(x)=``d/(dx)(x^3-9x^2+24x+2)`

`=d/(dx)(x^3)-d/(dx)(9x^2)+d/(dx)(24x)+d/(dx)(2)`

`=3x^2-18x+24+0`

`=3x^2-18x+24`

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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^2-18x+24 = 0`

`=>3(x^2-6x+8) = 0`

`=>3(x^2-2x-4x+8) = 0`

`=>3(x(x-2)-4(x-2)) = 0`

`=>3(x-2)(x-4) = 0`

`=>(x-2) = 0" or "(x-4) = 0`

`=>x = 2" or "x = 4`

The solution is
`x = 2,x = 4`

`:.` `x=2` and `x=4`

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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,2),(2,4),(4,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=1`

`f^'(1)``=3*1^2-18*1+24`

`=3-18+24`

`=9`` > 0`

2. For second interval `(2,4)`, we choose `x=3`

`f^'(3)``=3*3^2-18*3+24`

`=27-54+24`

`=-3`` < 0`

3. For third interval `(4,oo)`, we choose `x=5`

`f^'(5)``=3*5^2-18*5+24`

`=75-90+24`

`=9`` > 0`

Interval	x-value	`f^'(x)`	Negative or Positive
`(-oo,2)`	`1`	`f^'(1)=9`` > 0`	Positive
`(2,4)`	`3`	`f^'(3)=-3`` < 0`	Negative
`(4,oo)`	`5`	`f^'(5)=9`` > 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 maximum

At `x=4`, `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)``=2^3-9*2^2+24*2+2`

`=8-36+48+2`

`=22`

Local Maximum point = `(2,22)`

2. At `x=4`

`f(4)``=4^3-9*4^2+24*4+2`

`=64-144+96+2`

`=18`

Local Minimum point = `(4,18)`

graph

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