# What are the values and types of the critical points, if any, of #f(x)=x^4 - 8x^3 + 18x^2 - 27#?

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since the derivative has an odd order we have an inflection point.

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To find the critical points of ( f(x) = x^4 - 8x^3 + 18x^2 - 27 ), we first find its derivative and then solve for ( f'(x) = 0 ) to locate the critical points.

( f'(x) = 4x^3 - 24x^2 + 36x )

Setting ( f'(x) = 0 ), we solve for ( x ):

( 4x^3 - 24x^2 + 36x = 0 )

Factoring out ( 4x ), we get:

( 4x(x^2 - 6x + 9) = 0 )

( 4x(x - 3)^2 = 0 )

( x = 0 ) or ( x = 3 ) (repeated root)

Now, we find the corresponding ( y ) values:

When ( x = 0 ), ( f(0) = -27 ).

When ( x = 3 ), ( f(3) = 3^4 - 8(3)^3 + 18(3)^2 - 27 = -54 ).

So, the critical points of ( f(x) ) are ( (0, -27) ) and ( (3, -54) ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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