# Homework Help: I don't buy that it's possible for a function to not have critical points

1. Aug 29, 2010

### GreenPrint

1. The problem statement, all variables and given/known data

I don't believe that it's possible for a function to not have a local maxes and minimums. I have always been told that if you set the first derivative of a function to zero and solve for x and get imaginary numbers for your critical values that there are none. Now I don't by this because you can just keep on solving and get very defined values for critical points that are in the set of complex numbers... can maybe someone explain to me why it's considered not to have any critical points when you get complex numbers because by the looks of it every function has them they are just complex... like you can't deny that you can go through the same process to solve for critical points for equations that have them for equations that "don't" to get very defined values that are just complex... ? If they aren't critical points then what are they???

Just thinking about it a function MUST have a point were it turns around because if it didn't then it would just go on forever in one direction and never come back down...

For example the equation f(x) = (3X-1)/(2x^2 + x - 6) MUST have critical points were the function "turns around" because if it didn't it wouldn't be able to go back down... like when
f(0) = 1/6 f(1/2) = -.1 the function is decreasing and eventually becomes negative for values greater than 1/3... hmm so if there were no critical values then wouldn't you agree the function would just keep on going, becoming more and more negative as x becomes greater than 1/3... hmm but this isn't true as clearly it does turn around some were and become positive, for example f(3) = 8/15... how could f(x) become positive if there was no point were it turned around if it was just suppose to become more and more negative for values greater than 1/3???

For example setting the first derivative for this function equal to zero
x = 1/6 (2-7 i sqrt(2))
x = 1/6 (2+7 i sqrt(2))
so they are complex so there are "no" critical points... I'm not buying it... It seems as if these critical points are points for f(x) = x/0... now why is this? These seem to be the turning points... f(-2) and f(3/2) the y asymptote... hmmm

it's interesting that when you solve the first derivative when setting it equal to zero you only set the numerator and ignore the denominator completely because its "can't be defined", which I don't buy either but that is another issue...

sorry I'm just fascinated that people tell me that that f(x) = (3X-1)/(2x^2 + x - 6) has no critical points but yet it would appear that you can solve perfectly fine and get very well defined points... hope it makes sense can somebody explain why "there are none" as I'm not buying it... just thinking about it from a very logical perspective that there MUST be critical points to go from becoming more and more negative, if it's not suppose to turn around anywhere, but yet it becomes positive...? If there weren't any critical points wouldn't it just stay negative? hmmm....

Thanks!
Please tell me the truth please =)... no lies like cos(x)=-2 has no solutions lol...

2. Relevant equations

3. The attempt at a solution

Last edited: Aug 29, 2010
2. Aug 29, 2010

### rootX

f(x) = 2 is a function.

Some functions cannot be defined at some places, so they don't need to turn down or up, they can have different value at left side and different on the right side. These functions are not continuous.

3. Aug 29, 2010

### GreenPrint

Ok fine what about my example I used above =) Guess your right but what about f(x) = (3X-1)/(2x^2 + x - 6) hmm?

4. Aug 29, 2010

### GreenPrint

For equations with a variable =)

5. Aug 29, 2010

### Hurkyl

Staff Emeritus
For the real-valued function of the reals less the points 3/2 and -2,
f(x) = (3x-1)/(2x^2 + x - 6) ​
the point 1/6 (2-7 i sqrt(2)) is not a root of f'(x). f' is not even defined there, since it's not part of the domain!

This isn't logic, this is heuristics. :tongue: Heuristics that don't even apply, because, for example, the function does not vary continuously from 1/2 through 3.

The domain of your f is three disjoint intervals. In the first, it is strictly decreasing from 0 to $-\infty$ exclusive. In the second, it is strictly decreasing from $+\infty$ to $-\infty$ exclusive. In the third, it is strictly decreasing from $+\infty$ to 0 exclusive. Have you even tried plotting it?

The real-valued function cos of the real numbers does not have any points where it equals -2. The complex version of cos does, but that's a different function!

6. Aug 29, 2010

### Hurkyl

Staff Emeritus
The unary function f given by
f(x) = 2​
is a "function with a variable". It's just a constant function.

The functions defined by
f(x) = x​
similarly has no local minima or maxima. The function arctan as well.

7. Aug 29, 2010

### GreenPrint

then how come I get an answer =O

8. Aug 29, 2010

When we speak of critical points for a function such as

$$f(x) = \frac{3x-1}{2x^2+x-6}$$

the generally unspoken restriction (but still a restriction) is that the critical values should be real values. The derivative of your function is

$$f'(x) = -{{6\,x^2-4\,x+17}\over{\left(x+2\right)^2\,\left(2\,x-3\right)^2}}$$

There is no real value of x that solves $f'(x) = 0$. There are two values that leave $f'(x)$ undefined, so that the two critical values are $x = -2$ and $x = 3/2$.

If you refer to a critical point , then

* Some people use the terms critical point and critical value interchangeably: this can be confusing and, in my opinion, is an unfortunate choice. However, it's long established and I am not an arbiter of mathematical language usage
* If you think of a critical point as a point on the graph that may correspond to a max or a min, then (this is important) there must be a point on the graph of the function . If the function is not defined at a critical value, there cannot be a corresponding critical point.

So, as noted above, your function has two critical values, but since the function itself is not defined at those values, the function does not have any critical points.

9. Aug 29, 2010

### GreenPrint

also cos(x)=-2 has no solutions for reals yes but cos(x)=-2 has solutions when x is in the set of complex numbers =O

if your just given cos(x)=-2 and are not told anything, like if it's in the set of complex numbers or not or anything, then you can solve =O

10. Aug 30, 2010

### Staff: Mentor

But in most problems it should be understood what the domain is; in particular, whether the domain is the set of real numbers (or some subset of them), or it is the complex numbers.

If the problem involved finding the critical numbers of a real-valued function defined on the reals, you should not go looking amongst the complex numbers for the critical numbers.

11. Aug 30, 2010

### vela

Staff Emeritus
When you define a function, it's not enough to simply say, for instance, f(x)=2x. You need to specify the domain and codomain of the function as well. The function f:[0,1]→R, f(x)=2x, is different than the function g:R→R, g(x)=2x. The function f(x)=10 has no solution because x=5 is not in the domain of f whereas g(x)=10 does because its domain does include the point x=5. Similarly, for the function cos x:R→R, there is no solution to cos x=2 whereas for cos z:C→C, there is a solution to cos z=2.

As others have repeatedly noted in several threads now, the domain and codomain are often implied by the context, and you're just given some rule like f(x)=cos x. When you claim that "cos x=2 has no solutions" is a lie, either you're confusing different functions (e.g. reals to reals vs. complex to complex) or you're willfully ignoring the context of the problem. (To be fair, the latter may not be willful but rather due to poor reading comprehension, but in either case, it's not flattering.)