`f(x)=x^3-9x^2+24x+2`
Find Local maxima and minima

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`

`:.` The critical points are `x=2` and `x=4`

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Step-3: Apply the second derivative test
Now, `f^('')(x)=``d/(dx)(3x^2-18x+24)`

`=d/(dx)(3x^2)-d/(dx)(18x)+d/(dx)(24)`

`=6x-18+0`

`=6x-18`

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

For `x=2`

`f^('')(2)``=6*2-18`

`=12-18`

`=-6`` < 0`

`:.` At `x=2` the function is local maximum

For `x=4`

`f^('')(4)``=6*4-18`

`=24-18`

`=6`` > 0`

`:.` At `x=4` 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)=x^3-9x^2+24x+2`

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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