1

In class we were asked to find the range of

$$g (x)=\arcsin x+\arccos x+\arctan x$$

So to find minimum and maximum, I differentiated the function. $$g^\prime(x)=\frac{1}{1+x^2}$$ Now its maximum value is $1$, so the maximum of $g(x)$ is $\dfrac{3\pi}{4}$, but I couldn't get minima.

Also note that we can't take infinity, as $\arcsin$ and $\arccos$ are involved, which take values from $-1$ to $1$. Why don't we get a minimum?

Using inverse identities, I know the answer is $\left(\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right)$.

Quanto
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Archis Welankar
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  • Use https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values and http://math.stackexchange.com/questions/1116974/why-its-true-arcsinx-arccosx-frac-pi2 – lab bhattacharjee Oct 26 '16 at 06:52

3 Answers3

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The maximum value of the derivative is $1$, but this doesn't help in finding the maximum and the minimum of the function. What you get from the derivative is that the function is increasing on its domain, which is $[-1,1]$.

So, where and what are the minimum and maximum values?

egreg
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2

You have a strictly increasing function on a bounded interval. Therefore your minimum is at the beginning of the interval and thee maximum at the end. $$ min = g(-1) = \frac{\pi}{4}$$ $$ max = g(1) = \frac{3 \pi}{4}$$

BigPanda
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We have

$\arcsin (x)+\arccos (x)= \frac{\pi}{2}$ for all $x \in [-1,1]$.

Fred
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