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According to Weierstrass's8.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 )

*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.