How do you find the critical points for #f(x)=x^3-2x^2+3x#?

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Answer 1 · 2017-06-28T00:38:08

#x=1/3(2-isqrt5) and x=1/3(2+isqrt5)#

To find the critical points you first take the derivative using the power:

#d/dx=x^n=nx^(n-1)#

The derivative of #f(x)=x^3-2x^2+3x# is:

#d/dx=3x^2-4x+3#

Now you set it equal to zero and solve:

#3x^2-4x+3=0#

When you solve it, it will yield complex roots:

#x=1/3(2-isqrt5) and x=1/3(2+isqrt5)#

These two solutions are your critical numbers.