Confusion regarding interval on which a function is increasing
The question is as follows:
If the function $f(x)=\cos x$ is strictly increasing on the open interval $(0,\pi)$, where will it be increasing ?
The answer to this question is $[0,\pi]$.
I am a little confused here, $0$ and $\pi$ are the points where the derivative of $f(x)=\cos x$ is zero, so how can we conclude that the function will be increasing at these two points ? can't it be decreasing ?
$\endgroup$ 12 Answers
$\begingroup$Strictly increasing means that when some values $x_1 \lt x_2$ then $f(x_1) \lt f(x_2)$, that is to say the endpoints are not allowed to be equal. A function that is merely increasing means that $x_1 \lt x_2$ then $f(x_1) \le f(x_2)$, so the endpoints are allowed to be equal.
Since as you said $f^\prime(0) = f^\prime(\pi) = 0$ we can't say that the interval including the endpoints is strictly increasing, thus the open interval $(0,\pi)$ is the interval where $f(x)$ is strictly increasing and the closed interval $[0,\pi]$ is the interval where $f(x)$ is increasing non-strictly.
$\endgroup$ $\begingroup$A function is increasing if $f'(x) \geq 0$ and is strictly increasing if $f'(x) > 0$. So as you noted the difference is the inclusion of points whose derivative is zero.
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