# A silly problem on critical points?

A critical point or stationary point of a differentiable function of a single real variable, $f(x)$, is a value $x_0$ in the domain of $f$ where its derivative is $0$.

Until a few seconds ago, I thought that having a function $f(x)$ and It's derivative $f'(x)$, I'd just need to find points of value zero, that is $f'(x)=0$. But I've tried to do an exercise and find the critical points of $\cos x + 2x$, then having it's derivative: $-\sin x+2$ I've obtained a problem:

The codomain of $(-\sin x)$ is $[-1,1]$, if I add $2$, the codomain becomes $[1,3]$. This might be really silly, but how is it possible that $f'(x)=0$? No points of $f'(x)$ pass through zero.

Reading further:

A critical value is the image under f of a critical point. These concepts may be visualized through the graph of $f$ at a critical point, the graph has a horizontal tangent and the derivative of the function is zero.

And seeing the graph of $(-\sin x+2)$, I see that there are points in which the slope is zero, but none of them ever touch $y=0$. Is evaluating $f'(x)=0$ necessarily different of finding the critical points?

EDIT: From the second quote I've given, in the critical point, the graph has a horizontal tangent and the derivative of the function is zero. The first requirement is made, but Wolfram Alpha actually says that there are critical points: • Hi Jesus. This just shows that the function $\cos x + 2x$ has no critical points. – user99914 May 17 '15 at 6:42
• @John See here. – Billy Rubina May 17 '15 at 6:44
• I don't understand what is $\sin^{-1}(2)$..... – user99914 May 17 '15 at 6:46
• @John From the page: "$Sin^{-1}(x)$ is the inverse sine function." – Billy Rubina May 17 '15 at 6:49
• $\sin x$ is never $2$, so I do not know what is $\sin^{-1}(2)$. – user99914 May 17 '15 at 6:52