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my question is more of a conceptual one, but i'll use the problem i'm stuck on to keep things clear. I am confused about how to demonstrate whether a function is strictly monotonically increasing or decreasing etc. (i'm using the wrong brackets because the curly ones keep disappearing)

I have the function $$(f : x \in \mathbb{R} : x < 0) \rightarrow \mathbb{R}, f(x) = \frac{1}{x^{2}}$$

and I need to decide whether it is (strictly) monotonically increasing (or decreasing) and then show algebraically why this is the case.

I can see that it is strictly monotonically increasing and that it fits the inequality $$f(x_{1}) < f(x_{2})$$ for all $$x_{1}, x_{2} \in (-\infty, 0)$$ with $$x_{1} < x_{2}$$

but I am confused about how I show this algebraically. I'd really appreciate a general response to this that I can apply to similar problems.

Thank you very much.

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  • $\begingroup$ If $x_1<x_2<0$, then $-x_1>-x_2>0$ and hence $(-x_1)^2>(-x_2)^2>0$ or$x_1^2>x_2^2>0$. $\endgroup$ – Yiorgos S. Smyrlis Oct 12 '14 at 14:06
  • $\begingroup$ Thanks. Is it correct at the end to divide by $$x_{1}^{2}$$ and $$x_{2}^{2}$$ because this gives an inequality with the actual function? $\endgroup$ – jm22b Oct 12 '14 at 14:14
  • $\begingroup$ Yes, it is correct, since it is as if you multiply both terms of the inequality by positive numbers. $\endgroup$ – Yiorgos S. Smyrlis Oct 12 '14 at 14:16
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You can see that your function is monotonically for $x<0$ increasing here:

http://www.wolframalpha.com/input/?i=1%2F%28x%5E2%29

For the left (right) side of $0$, you can show strictly increasing (decreasing) by the sign of the derivative:

$$f(x)'=\frac{-2}{x^3}$$

So $f$ is strictly increasing for $x<0$ and strictly decreasing for $x>0$.

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  • $\begingroup$ We have just covered using the derivative for these types of questions in class, but in this case I needed to do it algebraically. Thanks all the same though, I think I have it cracked. $\endgroup$ – jm22b Oct 12 '14 at 14:15
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Hint: Just calculate $f'(x)$ see whether it is positive or negative or nonnegative or nonpositive in the domain .

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The answer lies in the derivative, $f'(x)=-\frac{2}{x^3}\gt 0,\,\forall x\lt 0$ thus the function is monotonic increasing.

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