Find the minimum value of solutions of an equation in the interval $(0,2)$ The equation is: $$x^{10}-10x^2+5 = 0$$
I know I have to use the mean value theorem, but I don't know how to apply it in this case.
 A: Instead of using the mean value theorem, looking graphically at the function $x^{10}-10x^2+5$ leads you to the conclusion that the turning point which can be found in the interval $(0,2)$ is the minimum of the same interval. This means your solution is:
$$\frac{d}{dx}(x^{10}-10x^2+5)=0$$
$$\Rightarrow 10x^9-20x=0$$
$$\Rightarrow 10x(x^8-2)=0$$
Therefore, either $10x$ is $0$ or $x^8-2$ is $0$. If $10x$ is $0$, then $x=0$ is one of the turning points. However, $0\notin(0,1)$. THerefore $x^8-2=0$ and thus the x coordinate of the turning point you need is $2^{1/8}\approx{1.09050773
…}\in(0,1)$. Putting this into your equation leads to the value $-8\times2^{1/4}+5\approx{4.51365692…}$, your answer.
A: Take f(x) = x¹⁰-10x²+5, f is continuous on (0,2). f'(x)=10x(x⁸-2), now here we use monotonous property of a function. Now, f'(x)=10x(x-2⅛)(x+2⅛)(x²+2¼)(x⁴+2½).
Now clearly on (0,2⅛), f'$\le$0, f is decreasing on (0,2⅛), and on (2⅛,2), f'$\ge$0, f is increasing on (2⅛,2).
So, clearly 2⅛ is the only point in (0,2) , where f takes it's minimum by monotonous property.
