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Consider the following function

$$f(x) = x^{\frac{1}{3}} - 9$$

How can I find the open interval on which the function is increasing or decreasing?

a. I know the critical points are $(0,-9)$

b. I know that the interval is not decreasing.

c. I know that by applying first derivative test to identify the relative extremum. (In this case no extrema).

All I cannot figure out is: in which interval the function is increasing?

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  • $\begingroup$ What do you mean with the critical points being $(0,-9)$? $\endgroup$ – Git Gud Jun 27 '13 at 17:58
  • $\begingroup$ Wrong typing: f(x)=x^1/3 -9 $\endgroup$ – Sara Sharp Jun 27 '13 at 18:01
  • $\begingroup$ That's what it currently is: $\sqrt[3]x=x^{1/3}$. $\endgroup$ – Git Gud Jun 27 '13 at 18:02
  • $\begingroup$ Did you mean $\displaystyle x^{(1/3)-9}$? If so, you should have used parentheses. $\endgroup$ – Git Gud Jun 27 '13 at 18:05
  • $\begingroup$ In what intervals is $f'(x)\ge0$? Note $f$ is not differentiable at $x=0$, but $f$ is continuous there (I'm assuming you want $f(x)=x^{1/3}-9$). This should be all you need... $\endgroup$ – David Mitra Jun 27 '13 at 18:09
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To find where a function is increasing or decreasing, we look at the first derivative: $$f(x) = x^{1/3} - 9\\ f'(x) = \frac{x^{-2/3}}{3}$$

We set the derivative equal to zero: $$\frac{x^{-2/3}}{3} = 0$$ $$x^{-2/3} = 0$$ $$\frac{1}{x} = 0$$

Critical points come in two categories: stationary and singular. Stationary points are where the first derivative is equal to zero; singular points are where it is undefined. Note that $\frac{1}{x} \ne 0$ for any $x$; thus, there are no stationary points!
However, the function $f'(x)$ is undefined at $x=0$, so there is a singular point at $x=0$.

Now, we test $f'(x)$ to see if it is positive or negative on either side of the origin: $$f(-1) = \frac{1}{3},\qquad f(1) = \frac{1}{3}$$

Thus, the first derivative is positive for all x values, except $x=0$ (where it is undefined). This tells us that the function is increasing on the interval $(-\infty, 0)$ and on $(0, \infty)$. But what about at $x=0$?

As $f$ is continuous at $x=0$, $f$ is increasing on all of $\mathbb{R}$. So, our final interval is $(-\infty, \infty)$.

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  • $\begingroup$ Thank you. My first answer was Increasing at (0,Infinity) which was wrong. $\endgroup$ – Sara Sharp Jun 27 '13 at 18:15
  • $\begingroup$ And (-Infinity,0) is wrong to. $\endgroup$ – Sara Sharp Jun 27 '13 at 18:16
  • $\begingroup$ yes that's correct thanks $\endgroup$ – Sara Sharp Jun 27 '13 at 18:20
  • $\begingroup$ Well I think every software has deferent ways for the answer. $\endgroup$ – Sara Sharp Jun 27 '13 at 18:31
  • $\begingroup$ Very nice now. +1 (Sorry to have nitpicked.) $\endgroup$ – David Mitra Jun 27 '13 at 18:44
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Sara, your answer $(-\infty , 0)$ is not wrong, since it is true that the function is increasing in this interval. The same holds for $(0 , + \infty)$. The answer must be $(- \infty , + \infty )$ only if the question was to find the largest interval where the function is increasing or decreasing :)

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  • $\begingroup$ Thanks. (-infinity,+infinity) was a correct answer. $\endgroup$ – Sara Sharp Jun 28 '13 at 19:27

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