Challenge Math Challenge - October 2018

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Periwinkle

8. Let $f$, $g$: $\mathbb{R} \rightarrow \mathbb{R}$ be two functions with $f\,''(x) + f\,'(x)g(x) - f(x) = 0$. Show that if $f(a) = f(b) = 0$ then $f(x) = 0$ for all $x\in [a,b]$. (by @QuantumQuest )
According to Weierstrass's maximum value theorem, if a function is continuous on $[a,b]$, then it is bounded on this interval and takes its maximum and minimum. If function $f(x)$ is not constant, then its maximum or minimum is different from $f(a) = f(b)=0$. Let this be the maximum that the function takes at an internal $x_0$ point of the interval. According to one of Fermat's theorems, in this case - provided that the function $f (x)$ can be derived at this point - the derivative $f'(x_0)$ here is $0$. Therefore, the following equation is correct:
$$f\,''(x_0) + 0 \cdot g(x_0) - f(x_0) = 0$$ $f(x_0)$ is positive, so $f\,''(x_0)$ is also positive. Therefore, the function $f(x)$ has a local minimum at $x_0$, which cannot be a local maximum.

"Math Challenge - October 2018"

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