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

[Wi_N]

Homework Statement

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

Relevant Equations

Surely there is a law about this? Im thinking its a straight line in that case.

so ergo the function is neither convex nor concave. but graphing it in a machine it looks convex...

 

[WWGD]

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]

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?

 

[fresh_42]

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.

 

[Mark44]

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?

 

[Wi_N]

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

 

[Mark44]

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]

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.

 

[RPinPA]

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 $>$.

 

[Wi_N]

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.

 

[Mark44]

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 $x \to -\infty$.

 

[Wi_N]

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.

 

[Mark44]

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?

 

[Wi_N]

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.

 

[Wi_N]

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??

 

[lekh2003]

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.

 

[Wi_N]

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.

 

[lekh2003]

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.

 

[Mark44]

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).

 

[Wi_N]

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.

 

[Mark44]

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.

 

[Wi_N]

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?

 

[Mark44]

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.

 

[Wi_N]

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?

 

[Mark44]

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?

For your problem, if x > -2, f(x) = -x, so the graph is actually linear on this interval.

 

[Wi_N]

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?

 

[Mark44]

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.

 

[Wi_N]

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.

 

[Mark44]

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.