How can I numerically find the local maximum values of a function? $y=1000 + (100/6)xe^{\sin x}$. Interval $[0 - 100]$. I tried differentiating it but I get stuck at solving the differential being equal to zero. Is there a way to find this?
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1Local maximums? Within what interval? If you take a series $x_n=\frac{\pi}2+2\pi n$... – abiessu Oct 20 '22 at 12:50
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Sure! Local maximums. Can you please give a longer version of the series? And any info on how to get it? Is it a taylor series? – Lemour Oct 20 '22 at 12:57
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That series is set for the maximums of $\sin x$ which will allow for starting points for the function using a numerical method. It's just the points at which $\sin x=1$. – abiessu Oct 20 '22 at 16:27
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The function is not bounded above. If you take $f(x) = 1000 + 100xe^{\sin(x)}$ and consider the sequence $(x_n) = 2\pi n$, you have $f(x_n) = 1000 + 100 \times 2\pi n \times e^0 = 1000 + 200\pi n$, which clearly diverges to $+\infty$.
The function is also not bounded below, using a similar reasoning with $(x_n) = -2\pi n$
For these reasons, there is no "finding maximum values" for this function.
As for local maximums, you have to study the zeroes of $f'(x)$, which are a necessary condition for extremums to be present.
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Then you need to look for the zeroes of $f'(x)$. We have $f'(x) = 100e^{\sin(x)}(1 + x\cos(x))$, hence you need to look for the zeroes of $1 + x\cos(x)$ – Gautier Blandin Oct 20 '22 at 13:21
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