# Thread: Simple problem decreasing function

1. ## Simple problem decreasing function

Can someone tell me the answer to this question?
I cannot seem to figure it out

The function $\displaystyle y=\frac{2}{x}$ is decreasing in??

A.(0,$\displaystyle \infty$)

b.($\displaystyle -\infty$,0)

c.(0,2)

d,($\displaystyle -\infty$,$\displaystyle \infty$)

I think B is the correct option but a lot of other people say it's wrong.

2. ## Re: Simple problem decreasing function

You posted this in the Calculus section so I would imagine you know how to take a derivative and how to use it. A differentiable function is decreasing if and only if its derivative is negative. What is the derivative of $\displaystyle y= 2x^{-1}$?

3. ## Re: Simple problem decreasing function

Well it's derivative would be $\displaystyle \frac{dy}{dx}=-\frac{2}{x^{2}}$.

4. ## Re: Simple problem decreasing function

I do not like this problem, at least not if you posted it correctly, because multiple choice questions normally make an implicit assumption, which is false in this case.

Before getting to the final answer, what is the domain of this function? You cannot possibly know where it is increasing or decreasing if you don't know where it even exists. It's what the lawyers call a threshold question.

You properly computed the derivative. The derivative is a function too. What is the derivative's domain? Where is the derivative positive? Where is the derivative negative? So where is the parent function decreasing?

5. ## Re: Simple problem decreasing function

So for what values of x is that negative? As JeffM pointed out, there is a problem with this question- NONE of the given choices is correct!

6. ## Re: Simple problem decreasing function

The domain of a function, if not specified, is assumed to be all real numbers for which the formula makes sense. So in this case $\displaystyle (-\infty,0)\cup(0,\infty)$.

I think the definition of decreasing (at a point) ends up being equivalent to the derivative existing and being less than zero (if "decreasing" means strictly decreasing) or less than or equal to zero (if "decreasing" means non-increasing). There's no difference for this function.

The function $\displaystyle y=\frac{2}{x}$ is decreasing at every point in its domain, so the best answer to the question is $\displaystyle (-\infty,0)\cup(0,\infty)$.

The function is decreasing on $\displaystyle (-\infty,0)$, $\displaystyle (0,\infty)$, and $\displaystyle (0,2)$ since they are subsets of $\displaystyle (-\infty,0)\cup(0,\infty)$. Since it is not defined at 0, the answer $\displaystyle (-\infty,\infty)$ is incorrect. So my opinion is that A, B, and C are all correct (the function is decreasing on those intervals), and D is incorrect.

- Hollywood

7. ## Re: Simple problem decreasing function

Hollywood is of course technically correct, but the implication in a multiple choice question is that only one answer is correct. That was the point that Halls of Ivy and I were trying to make: the student was being psychologically misled by the form of the question.

A well structured question might have been A and B as stated.

C. Both A and B are correct.

D. Only B is correct.

Such a problem would have taught the student something. Bad problems can seriously impede the learning process.

EDIT: Upon reflection, it is possible that option D is not how the original question was worded. I really should not criticize the problem's author without the qualification that I am assuming the OP transcribed the problem completely and exactly, an assumption that is far from certain

8. ## Re: Simple problem decreasing function

Originally Posted by HallsofIvy
So for what values of x is that negative? As JeffM pointed out, there is a problem with this question- NONE of the given choices is correct!
On looking this over, it occurs to me that this problem does NOT ask for the entire set on which the function is decreasing. It only asks which of the given sets has the property that the function is decreasing on it. Yes, the derivative of this function is negative for x for which it is defined. Every choice except (d) is correct since (d) is the only one that contains 0.