Please show me how to approach and solve the problem: Find the minimum value of $ f(x)=x^{2}-4x+3+3\cos(\frac{\pi x}{4})$

Question

Problem: Find the minimum value of $ f(x)=x^{2}-4x+3+3\cos\left(\frac{\pi x}{4}\right)$

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At first, I found the minimum value of each functions $g(x)=x^{2}-4x+3$ and $h(x)=3\cos\left(\frac{\pi x}{4}\right)$ but they do not happen at the same value of $x$.

Then, I use the derivation of $f(x)$ to find the exact value of $x$. $f'(x)=2x-4-3\frac{\pi }{4}\sin\left(\frac{\pi x}{4}\right)$.
At this step, i found it difficult to solve it because of the complicated function.
Although I mean to use the graph of the derivation, I think it is still hard.

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Please help me to find some approaches that are suitable to high school's curricula. Thank you!

Answer 1

The derivative of the function is

$$2x-4-\frac{3\pi}4\sin\frac{\pi x}4.$$

This is indeed a transcendental function for which there is no "easy" root, and you can bet that that it cannot be solved analytically.

Numerically, $x\approx2.8973954310014$ is the only solution, corresponding to a minimum.

Answer 2

Using the derivative, by inspection or plotting, we can see that the solution is close to $x=3$.

So, make a Taylor expansion to get for $$y=2 x-\frac{3}{4} \pi \sin \left(\frac{\pi x}{4}\right)-4$$ $$y=\left(2-\frac{3 \pi }{4 \sqrt{2}}\right)+\left(2+\frac{3 \pi ^2}{16 \sqrt{2}}\right) (x-3)+\frac{3 \pi ^3 }{128 \sqrt{2}}(x-3)^2+O\left((x-3)^3\right)$$ Now, using series reversion to obtain $$x=3+t-\frac{3 \pi ^3 }{256 \sqrt{2}+24 \pi ^2}t^2+O\left(t^5\right)\qquad \text{where} \qquad t=\frac{4 \left(8 (y-2)+3 \sqrt{2} \pi \right)}{64+3 \sqrt{2} \pi ^2}$$ Making $y=0$, an approximation is $x=2.897492$ while the exact solution, given by @Yves Daoust, is $x=2.897395$.

Edit

We can have a quite good approxiamtion of the solution using the approximation $$\sin(t) \simeq \frac{16 (\pi -t) t}{5 \pi ^2-4 (\pi -t) t}\qquad (0\leq t\leq\pi)$$ was proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.

Using it, after simplifications, we end with a cubic equation $$2 x^3-3(4- \pi ) x^2+4(14-3 \pi ) x-80=0$$ the real solution of which being $$x=2+\sqrt{\frac{64}{3}-\pi ^2} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{3 \sqrt{3} \pi \left(56-\pi ^2\right)}{\left(64-3 \pi ^2\right)^{3/2}}\right)\right)-\frac{\pi }{2}\approx 2.89641$$ This vaue would give a minimum value of $f(x)$ equal to $-2.13835345$ while a rigorous calculations lead to $-2.13835500$