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$f(x) = \left\{\begin{array}{lcl}-x^3 + 1&\text{if}&x\geq0\\ -x^2+2x&\text{if}&x<0\end{array}\right.$ I want to find the critical points but after differentializing and equalizing to $0$, the $x$ which one is 1 and other is 0. (I think that the point should be the same one?)

What's wrong with it?

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    @DonAntonio Yes, it seems quite widespread in calculus textbooks, at least in the US. But Spivak (Calculus, chapter 11) says that a point is critical if $f'(x)=0$. Here we tend to classify points of non-differentiability, but wwe reserve the name *critical* or *stationary* to the zeroes of the first derivative.2012-12-25

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Hint: We can see that $x=0$ is a critical point for the function because $f'^+(0)=-3(0^2)\neq 2=-2(0)+2=f'^-(0) $

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    $\quad \ddot\smile\quad$2013-03-03
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If I understand the post correctly, you view $x=1$ as a critical point of $f(x)$. It isn't. Although certainly the derivative of $-x^2+2x$ is $0$ at $x=1$, the function $f(x)$ is $-x^2+2x$ for negative $x$ only.

As to whether $0$ should be called a critical point, that depends on the definition being used in your course. The usual convention in elementary North American texts would include points of discontinuity of the function, and points where the derivative does not exist.