show that the equation has a root in the interval $(0, \pi)$ $$e^{x/\pi} + \sin(x) = x^2$$
I have solved it with the Intermediate Value Theory (as below) but I don’t think that my justification is enough, and I don’t know how to do it in a different way?
My answer:
let: $f(x) = e^{x/\pi} + \sin(x) - x^2$
Note that: $f(0) = 1+0-0 = 1$
Note that: $f(\pi) = e+0-\pi^2 =$ a negative number (how can I justify this w/o a calculator)
So, by the intermediate value theory, all numbers between $0$ and (that negative number) is defined.
Also, later I noticed that the interval in question is an open interval and it didn’t say that this equation is continuous. So, back to square one?
If you know $0<e<3<\pi$ that's enough to conclude $e-\pi^2<0.$
Let's look at how to show $3<\pi.$
Inscribe a regular hexagon in a circle of diameter $1.$ A bit of elementary geometry shows the perimeter of the hexagon is $3.$ Therefore the circumference of the circle is $>3,$ so you have $\pi>3.$
Now let's look at how to show $e<3.$
Suppose you know $\displaystyle e^x = \sum_{n=0}^\infty \frac {x^n}{n!}.$ Then $$ e^{-1} = 1 - 1 + \frac 1 2 - \frac 1 6 + \frac 1 {24} - \cdots. $$ Since the terms alternate in signs while decreasing, you must have $$ \frac 1 2 - \frac 1 6 < e^{-1} < \frac 1 2 - \frac 1 6 + \frac 1 {24}, $$ and therefore $e^{-1}>\dfrac 1 3,$ so $e<3.$