# If f''(x)=0, how do you find the convexity of the function?

• Wi_N
In summary, the conversation discusses a function that is neither convex nor concave. The function has a vertical asymptote within a certain range, making it impossible for the function to be a straight line. The conversation also touches on the definition of convex and concave, and the relationship between the second derivative and curvature. Ultimately, the conversation concludes that a function that hugs an asymptote cannot be a straight line.
Wi_N
Homework Statement
If f''(x)=0, how do you find the convexity of the function?
Relevant Equations
so ergo the function is neither convex nor concave. but graphing it in a machine it looks convex...

If you mean f''(x) is identically 0 for the function, then , yes, you can integrate and show it is a line, but if you meant f''(x)=0 at a particular point, it is a different issue. Which one did you mean?

Wi_N
WWGD said:
If you mean f''(x) is identically 0 for the function, then , yes, you can integrate and show it is a line, but if you meant f''(x)=0 at a particular point, it is a different issue. Which one did you mean?

i mean that i have a function that is f'(x) = c and f''(x)=0 within a certain range. however the function has a vertical asymptote within that specific range going to +infinity which makes it impossible for the function to be a straight line. what does the law say?

Have you calculated some examples? E.g. how does ##f(x)=x^3-3x^2+x+2## look like around ##x=1##, or the negative of oi ##f(x)=-x^3+3x^2-x-2##? And you are right, ##f(x)=mx+b## are also valid solutions.

Wi_N
Wi_N said:
i mean that i have a function that is f'(x) = c and f''(x)=0 within a certain range. however the function has a vertical asymptote within that specific range going to +infinity which makes it impossible for the function to be a straight line. what does the law say?
If f'(x) = c for all x in some interval, then there can't be a vertical asymptote in that interval. Also, since the derivative is defined on the whole interval, the function f has to be continuous on that interval, thereby ruling out the possibility of vertical asymptotes.

If the derivative is constant on some interval, the graph of the function on that interval is a straight line with slope c.

Wi_N
just making sure. (x^2 +2x)/(1-abs(x+3)) is a line after x>-3? cause f''(x)=0.
if the second derivative is 0 does that always mean the function is not concave nor convex?

Last edited:
ill ping this since i edited it. i still need help.

Wi_N said:
just making sure. (x^2 +2x)/(1-abs(x+3)) is a line after x>-3? cause f''(x)=0.
if the second derivative is 0 does that always mean the function is not concave nor convex?
This is the same function you asked about in another thread. If x > -2, (not -3) then ##f(x) = \frac{x^2 + 2x}{1 - |x + 3|} = \frac{x(x + 2)}{-x - 2} = -x##. So the graph of this function is a line with slope -1 on the interval ##(-2, \infty)##. A line has no curvature, so is neither concave up or concave down.

Wi_N
Mark44 said:
This is the same function you asked about in another thread. If x > -2, (not -3) then ##f(x) = \frac{x^2 + 2x}{1 - |x + 3|} = \frac{x(x + 2)}{-x - 2} = -x##. So the graph of this function is a line with slope -1 on the interval ##(-2, \infty)##. A line has no curvature, so is neither concave up or concave down.

what if f''(x)=c+ does that mean its convex? I am talking about x<-3 for this function.

Just one comment on this discussion: Depending on the author, the definition of "convex" and "concave" may include a straight line. Then a distinction is made between "convex" and "strictly convex", analogous to the distinction between ##>=## and ##>##.

Mark44 said:
This is the same function you asked about in another thread. If x > -2, (not -3) then ##f(x) = \frac{x^2 + 2x}{1 - |x + 3|} = \frac{x(x + 2)}{-x - 2} = -x##. So the graph of this function is a line with slope -1 on the interval ##(-2, \infty)##. A line has no curvature, so is neither concave up or concave down.

not sure why you're so adamant about x=-2 since the slope starts at -3 with a hole at x=-2.

Wi_N said:
what if f''(x)=c+ does that mean its convex? I am talking about x<-3 for this function.
But f''(x) is not a constant for the function here. In your other thread, I though we agreed that there was an oblique asymptote as ##x \to -\infty##.

Mark44 said:
But f''(x) is not a constant for the function here. In your other thread, I though we agreed that there was an oblique asymptote as ##x \to -\infty##.

take the second derivative of the function for when x<-3. we get 16/(x+4)^3

we had a y=-x asymptote for x>-3 when we did lim f(x)/x.

Wi_N said:
take the second derivative of the function for when x<-3. we get 16/(x+4)^3

we had a y=-x asymptote for x>-3 when we did lim f(x)/x.
First off, that's no a constant.
Second, I don't think that's correct for the second derivative.
What do you get for the first derivative for x < -3?

Mark44 said:
First off, that's no a constant.
Second, I don't think that's correct for the second derivative.
What do you get for the first derivative for x < -3?

first derivative is (x^2 +8x+8) / (x+4) ^2

its a constant for the fact that it would never be =0. OR WAIT mistake made here.
edit: its concave for x<-4.

the interesting point is what happens between -4 and -3 its a function that goes into infinity hugging an asymptote. could a function that is hugging a asymptote be a straight line??

It's physically impossible for there to exist an asymptote for any f such that f'(x) = c. A vertical asymptote is of slope ##\infty##, which is definitely not the value of f'(x).

Also I'm confused at what your question is at this point.

lekh2003 said:
It's physically impossible for there to exist an asymptote for any f such that f'(x) = c. A vertical asymptote is of slope ##\infty##, which is definitely not the value of f'(x).

Also I'm confused at what your question is at this point.

i have it figured out, thanks. just answer this question. can a function that huggs an asymptote from (a,b) be a straight line? the asymptote is a vertical line.

Wi_N said:
i have it figured out, thanks. just answer this question. can a function that huggs an asymptote from (a,b) be a straight line?
No, it cannot, this doesn't make graphical sense either. An asymptote is a line, and a function "hugs" the asymptote if it is approaching said line, but never touches it. Key word: approaching. It needs to be getting closer, without touching. This is impossible for a straight line function. The line would have to be at skew to the line of the asymptote, but then that would mean that they would intersect.

Wi_N said:
first derivative is (x^2 +8x+8) / (x+4) ^2

its a constant for the fact that it would never be =0. OR WAIT mistake made here.
edit: its concave for x<-4.
OK, it's not constant for x < -4 -- the 2nd derivative is negative for x < -4, but gets closer to 0 as ##x \to -\infty##. Since f''(x) < 0 for x < -4, the graph is concave down (the terminology I favor).

Mark44 said:
OK, it's not constant for x < -4 -- the 2nd derivative is negative for x < -4, but gets closer to 0 as ##x \to -\infty##. Since f''(x) < 0 for x < -4, the graph is concave down (the terminology I favor).

what is the function between -4 and -3? the maths says neither since f''(x)=0. but apparently a function that huggs a vertical line can't be a straight line either..

edit: nevermind its convex. thanks for all the help.

Wi_N said:
what is the function between -4 and -3? the maths says neither since f''(x)=0. but apparently a function that huggs a vertical line can't be a straight line either..
The function is undefined at x = -4.
Look at the the function's behavior near x = -4 to see what the graph is doing.
If x < - 3, ##f(x) = \frac{x^2 + 2x}{x + 4}##
If x < -4, the numerator is positive, and the denominator is negative. Can you see why?
If -4 < x < -3, the numerator is still positive, but what sign is the denominator?
Notice that the closer x is to -4, the fraction gets either a lot larger or a lot more negative.

Mark44 said:
The function is undefined at x = -4.
Look at the the function's behavior near x = -4 to see what the graph is doing.
If x < - 3, ##f(x) = \frac{x^2 + 2x}{x + 4}##
If x < -4, the numerator is positive, and the denominator is negative. Can you see why?
If -4 < x < -3, the numerator is still positive, but what sign is the denominator?
Notice that the closer x is to -4, the fraction gets either a lot larger or a lot more negative.

Yes thanks, i edited my reply. i have one last question. the function from x>-3 seems to be overlapping the asymptote at y=-x is that really an asymptote then or a straight line? is both possible?

Wi_N said:
Yes thanks, i edited my reply. i have one last question. the function from x>-3 seems to be overlapping the asymptote at y=-x is that really an asymptote then or a straight line? is both possible?
Horizontal or oblique asymptotes are relevant only for very large or very negative values of the independent variable. It doesn't make any sense to discuss a horizontal or oblique asymptote on some interval with finite endpoints.
If ##\lim_{x \to \infty} f(x) = \text{constant}## then the horizontal line y = constant is a horizontal asymptote. The situation is similar if the limit goes to negative infinity.
If If ##\lim_{x \to \infty} f(x) =kx##, then the line y = kx is an oblique asymptote, and similar if the limit goes to negative infinity.
For your problem, if x > -2, f(x) = -x, so the graph is actually linear on this interval. If x < -4, If ##\lim_{x \to -\infty} f(x) = x##, so the graph approaches the line y = x as x goes to negative infinity, but always lies below it.

Ill make my question easier:

Can y=-x be an asymptot to the function if that function is not convex nor concave for that particular intervall?

Wi_N said:
Ill make my question easier:

Can y=-x be an asymptot to the function if that function is not convex nor concave for that particular intervall?
No. If the second derivative is zero on an interval, the graph is a straight line on that interval.

Did you see what I wrote in my previous post?
Mark44 said:
For your problem, if x > -2, f(x) = -x, so the graph is actually linear on this interval.

Mark44 said:
No. If the second derivative is zero on an interval, the graph is a straight line on that interval.

Did you see what I wrote in my previous post?

then how come i got y=-x as an asymptot for that intervall?

Wi_N said:
then how come i got y=-x as an asymptot for that intervall?
I don't know how you got that.
If x > -2, the function simplifies to f(x) = -x. The graph isn't asymptotic to the line y = -x -- the graph is that line, for x > -2.

Mark44 said:
I don't know how you got that.
If x > -2, the function simplifies to f(x) = -x. The graph isn't asymptotic to the line y = -x -- the graph is that line, for x > -2.

lim fx/x => x^2 + 2x / (1-(x+3)x = (x^2+2x)/(-2x-x^2) = -(x^2 +2x)/(x^2 +2x)=-1
even the mentors of this forum told me it was an asymptote in the previous thread.

edit: it was infact you that told me it was.

Wi_N said:
lim fx/x => x^2 + 2x / (1-(x+3)x = (x^2+2x)/(-2x-x^2) = -(x^2 +2x)/(x^2 +2x)=-1
even the mentors of this forum told me it was an asymptote in the previous thread.

edit: it was infact you that told me it was.
If I said that y = -x was an oblique asymptote, it was because I hadn't worked through the problem far enough to recognize that f(x) = -x, if x > -2.

Saying that this function has an oblique asymptote of y = -x, for x > -2, is kind of like saying that f(x) = 2x has an oblique asymptote of y = 2x.

A graph that has an oblique or horizontal asymptote gets close to, but doesn't touch or cross the asymptote for large x for very negative x. So y = -x is not an oblique asymptote for the function in this thread.

Getting to your original question where ##f''(x) = 0## can the function be convex. Others have pointed out if you have this equation holding in an interval, the function is linear there. But if you just have ##f''(c)=0## for a single point, that just tells you that the function may be concave up, down, or neither depending on higher derivatives at that point. Think about ##y = x^4## at ##x=0##. It is obviously concave up at ##x=0## but ##f''(0) = f'''(0) = 0## and ## f^{4}(0)=24##. But ##x^3## has neither concavity at ##0##. And for ##x^n## for larger values of ##n## it's the same story, only more so.

archaic
If ##f''(c)=0## but doesn't change sign before and after, then the concavity also doesn't change. If it changes its sign, then the concavity changes and ##c## is called an inflection point.
A linear function is both concave up and down, though.

LCKurtz said:
or neither
How?
EDIT : Oh, you mean an inflection point.

It could literally be neither without being an inflection point. Perhaps the graph has a straight line segment at ##c##. Unless you want to consider a straight line to be both concave up and down, which I don't think is standard convention. Nor would you consider such a point an inflection point since there is not a change of concavity at ##c##.

## 1. What does it mean for a function to have a second derivative of zero?

Having a second derivative of zero means that the function's rate of change is neither increasing nor decreasing at that point. This can indicate a point of inflection, where the curvature of the function changes.

## 2. How do you determine the convexity of a function with a second derivative of zero?

To determine the convexity of a function with a second derivative of zero, you must look at the behavior of the first derivative. If the first derivative is increasing, the function is concave up and if it is decreasing, the function is concave down.

## 3. Can a function have a second derivative of zero and still be convex?

Yes, a function can have a second derivative of zero and still be convex. This can occur when the function has a constant slope, meaning it is neither concave nor convex.

## 4. How do you graph a function with a second derivative of zero?

To graph a function with a second derivative of zero, you can plot the points where the function has a second derivative of zero and then use the behavior of the first derivative to determine the shape of the graph.

## 5. Are there any real-world applications of a function with a second derivative of zero?

Yes, there are real-world applications of functions with a second derivative of zero. For example, in economics, a function with a second derivative of zero can represent a point of equilibrium where supply and demand are equal.

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