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.

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