The critical values of a curve f(x) are thepoints in its domain where the derivative f'(x) is zero or not defined.
Given:
It is given that
[tex]f(x)=x^3-3x^2+29[/tex]
To find critical points first let us find f'(x)
[tex]\begin{gathered} f^{\prime}(x)=\frac{d}{dx}(x^3-3x^2+29) \\ =3x^2-6x \\ =3x(x-2) \end{gathered}[/tex]
Now on setting f'(x)=0 we have,
[tex]\begin{gathered} 3x(x-2)=0 \\ \Rightarrow3x=0,x-2=0 \\ \Rightarrow x=0,2 \end{gathered}[/tex]
So, the critial values are x=0,2.
