`f(x) = x^3 - 3x^2 + 3` Find all relative extrema, use the second derivative test where applicable.

Textbook Question

Chapter 3, 3.4 - Problem 33 - Calculus of a Single Variable (10th Edition, Ron Larson).
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thilina-g | College Teacher | (Level 1) Educator

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First derivative to locate the x coordinates of critical points.

`f'(x) = 3*x^2-3*2*x+0 = 3x^2-6x `

At critical points, `f'(x) = 0,`

`3x^2-6x = 0`

`x^2-2x=0`

`x(x-2) =0`

This gives, `x = 0 or x =2`

Therefore, there are two critical points at x = 0  and x = 2.

Second derivative test.

`f''(x) = 6x -6`

At `x = 0, f"(x) = 6*0-6 = -6 lt0` . Therefore, at x =0, there is a local maximum.

at `x = 0, f(x) = 0^3-3*(0)^2+3 = 3` The local maximum is `(0,3)`

at `x=2, f''(x) = 6*2-6 = +6 gt3` Therefore, at x = 2, there is a local minimum

at `x = 2, f'(x) = 2^3-3*(2)^2+3 = 8-12+3 = -1` The local minimum is `(2,-1)`

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loves2learn | (Level 3) Salutatorian

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Find the first derivative.

Given,

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

Then,

`f'(x)=3x^2-6x`

Now we need to find the value(s) of x that make the first derivative zero to find the critical points.

`0=3x^2-6x`

`0=3x(x-2)`

`x=0` and `x=2`

These will be a relative extrema if it changes sign, so find 2 values around both to test using the first derivative test, like -1 and 1 for the first  and 1 and 3 for the second.

`f'(-1)=9`

`f'(1)=-3`

Therefore, this a relative extrema, and find the y-value for the x `(0,3)`

Do the same for the other critical point

`f(1)=-3`

`f'(3)=9`

This is also a relative extrema.

Find the y-value `(2,-1)`

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