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Is there any way to figure out the range of values of the function $$y=\frac{2}{x}\cdot \sin(x)?$$ The domain is so easy to know. It's all real numbers except $0$. However the challenging part is to figure out what is the range. Any ideas?

I know that I need to know the minimum & maximum values that the function $y$ can get but no idea how to figure that out. All I know is that the possible values of $\sin(x)$ alone is from $-1$ to $1$ but the complexity comes from that part $2/x$ which I don't know what to do when this part is multiplied by $\sin(x).$ Thanks in advance.

M47145
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M.Samuel
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3 Answers3

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You indicate that your function is $\frac {2 \sin x}x$. Since $\lim_{x \to 0} \frac {2 \sin x}x=2$ the upper end of the range is $2$, but it won't get there. To get the minimum, we take the derivative and set to zero: $0=\frac d{dx} 2\frac {\sin x}x=2\frac {x\cos x-\sin x}{x^2}$ or $x=\tan x$ You can't solve this algebraically, but can solve it numerically. Alpha gives an approximate solution of $4.4934$ witha a value of about $-0.4345$, so the range is $(\approx -0.4345,2)$

Ross Millikan
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Often it is quite difficult to know the range of functions. You can plot it and look, but that might be misleading if the plot is not representative.

vadim123
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Well, assuming that you mean $\frac{2}{x}\sin(x)$: note that for $x << 1$, $\sin (x) \sim x$, i.e, for $x = 0$ you have $y = 2$. Then, for $x \rightarrow \infty$, $y \rightarrow 0$. What is left is to look at the derivative around 0 to make sure that $x = 0$ is a local (and global) maximum

Alex
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