How to calculate the absolute extrema for this equation Basically, I am trying to find the absolute extrema for this problem, but I am stuck after a few steps.
$$y=\frac{4}{x}+\tan\left(\frac{\pi x}{8}\right),\quad x\in[1,2].$$
I have figured out the derivative, which is
$$y'=-\frac{4}{x^2}+\sec^2\left(\frac{\pi x}{8}\right)\cdot\frac{\pi}{8}.$$
However, after reaching this step, I am stuck. I have tried setting the derivative equal to zero, but I do not get anywhere as I don't know what to do next. How can I solve this problem? Any help is appreciated. Thanks.
 A: Hint: You can use a difference of squares with the derivative:
$$ \frac{\pi}{8} \sec^2 \left( \frac{\pi x}{8} \right) - \frac{4}{x^2} = 0 $$
$$ \Rightarrow \left(\sqrt{\frac{\pi}{8}} \sec \left( \frac{\pi x}{8} \right) + \frac{2}{x}\right) \left(\sqrt{\frac{\pi}{8}} \sec \left( \frac{\pi x}{8} \right) - \frac{2}{x}\right) = 0 $$
Update:
$$ \Rightarrow \pm \frac{2}{x} = \sqrt{\frac{\pi}{8}} \sec \left( \frac{\pi x}{8} \right)$$
$$ \Rightarrow \pm \frac{x}{4}\sqrt{\frac{\pi}{2}} = \cos \left( \frac{\pi x}{8} \right) $$
Let $f_1(x) = \frac{x}{4}\sqrt{\frac{\pi}{2}}$, $f_2(x) = -\frac{x}{4}\sqrt{\frac{\pi}{2}}$, and $g(x) = \cos \left( \frac{\pi x}{8} \right) $.
We'll start with $f_1(x) = g(x)$. First, note $f_1(0) = 0$, and $g(0) = 1$. The period of $g(x)$ is $16$, so the next time it reaches its maximum value of $1$ is at $x = 16$. At the same time, $f_1(x) = 1$ at $x = \frac{4\sqrt{2}}{\pi} < 16 $. Since $f_1$ strictly increasing, $f_1$ and $g$ intersect exactly once on $x\ge 0$, somewhere on $x \in \left[0,\frac{4\sqrt{2}}{\pi}\right] $.
For $x \le 0$, $f_1(x) = -1$ at $x = -\frac{4\sqrt{2}}{\pi} > -4 $. Again, since $f_1$ is strictly increasing, $f_1$ is below the minimum value of $g$ for all $x < -\frac{4\sqrt{2}}{\pi} $ . Since $g(x) > 0$ for $x \in (-4,0] $, we have $f_1(x) < g(x) $ for all $x \le 0$.
Combining these, we have $f_1$ and $g$ intersect exactly once, let's say at $x = r_1$, where $r_1 \in \left[0,\frac{4\sqrt{2}}{\pi}\right] $.
Then, since $f_2(x) = f_1(-x)$ and $g(x) = g(-x)$, $f_2$ and $g$ intersect exactly once at $x = r_2 = -r_1$.
Further, note that there must be a sign change in $y'$ at $r_1$ because $f_1$ is strictly increasing for all $x$, and $g$ is decreasing on $[0,8]$, implying that $g$ is decreasing at $r_1$. The same logic implies a sign change at $r_2$ as well.
These two solutions are the unique roots of the factored form of the derivative we had above.
Then, we have
$$ y'|_{x=1} = -4 + \frac{\pi}{8}\sec^2 \left( \frac{\pi}{8} \right) $$
$$ = -4 + \frac{\pi}{8}\frac{4}{2+\sqrt{2}} $$
$$ = -4 + \frac{\pi}{4+2\sqrt{2}} $$
$$ < 0 . $$
Similarly,
$$ y'|_{x=2} = -1 + \frac{\pi}{8}\sec^2 \left( \frac{\pi}{4} \right) $$
$$ = -4 + \frac{\pi}{8}*2 $$
$$ = -4 + \frac{\pi}{4} $$
$$ < 0 . $$
Since $y' < 0$ at both $x = 1$ and $x = 2$, we know that $r_1 \notin [1,2]$, since there is no sign change in $y'$ and there is exactly one root (which has a corresponding sign change) for $x \ge 0$.
Then, the extreme values of $y$ must be at $x = 1$ and $x = 2$.
A: You can use the inequality $\quad\cos(x)\ge 1-\dfrac{x^2}{2}$
Once factored $y'(x)=\dfrac{\pi x^2-32\cos(\frac{\pi x}8)^2}{8x^2\cos(\frac{\pi x}8)^2}$, and since the denominator is always positive in $[1,2]$, we can focus on the numerator.
We have $N(x)\le \pi x^2-32(1-\frac 12(\frac{\pi x}8)^2)^2=\pi x^2-32+\frac 12\pi^2 x^2-\frac 1{512}\pi^4 x^4$
This is a quadratic in $u=x^2$ and it can be solved classically (though a bit ugly...), in particular $N(x)\le 0$ between the roots $x\approx \pm 2.103$ which cover the interval of interest $[1,2]$.
So $y'<0$ on $[1,2]$ so the minimum is $y(2)$.
A: To continue after your last step . Simplifying after setting the derivative to zero,
$$ \dfrac{x}{2}=\cos \sqrt{\dfrac{\pi x}{8}}$$
Can be solved by iterative methods numerically, but no closed form.
A: Since you already received good answers, let me show something amazing.
With the derivative you wrote, you cannot find its zero except using numerical methods.
To simplify the problem, let $x=\frac{8 t}{\pi }$ and the function becomes
$$f(t)=\frac{\pi }{2 t}+\tan (t)$$
$$f'(t)=\sec ^2(t)-\frac{\pi }{2 t^2}$$ Now, use the $1,400$ years old approximation
$$\cos(t) \simeq\frac{\pi ^2-4t^2}{\pi ^2+t^2}\qquad \text{for} \qquad -\frac \pi 2 \leq x\leq\frac \pi 2$$  which makes
$$f'(t) \sim \frac{\left(\pi ^2+t^2\right)^2}{\left(\pi ^2-4 t^2\right)^2}-\frac{\pi }{2 t^2}$$ Let $t=\sqrt y$ and, after simplifications, the derivative write
$$\frac {2 y^3+4 (\pi -4) \pi  y^2+2 \pi ^3 (4+\pi ) y-\pi ^5 } {2 \left(\pi ^2-4 y\right)^2 y }$$ The cubic in numerator shows only one real root since
$$\Delta=-100 \pi^9 (8 \pi ^2+83 \pi-128) <0$$ Now, using the hyperbolic method, this root is
$$y=\frac{2\pi}3\Bigg[4-\pi+\sqrt{44\pi- \pi^2-64} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{2048-2112 \pi +411 \pi ^2+4 \pi ^3}{4(44\pi- \pi^2-64)^{3/2}}\right)\right)  \Bigg]$$ which is $y\sim 0.701420$; then $t\sim 0.837508$ and finally $x\sim 2.13270$ while the solution given by Newton method is $x=2.13448$.
Using the approximate solution, the minimum of the function would be $2.98561$ while its "exact" value is $2.98561$ !!
