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

Answer 1

Evelyn Agard

#x=3,f(3)=0# is an inflection point. and for #(0;-27)# a minimum point.

Your #f(x)# is equal to

#f(x)=(x+1)(x-3)^3#

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

#x=0# or #x=3#

#f''(x)=12x^2-48x+36#

#f''(0)=36>0# we get a minimum

#f''(3)=0#

#f'''(x)=24x-48#

#f'''(3)=48ne 0#

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

We have

#(0,-27)# a minimum

and for

#(3;0)# an inflection point.

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Answer 2

Paisley Johnson

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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Answer from HIX Tutor

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