Extreme difficulty with a concavity problem

[Solitare]

Homework Statement

1. Determine the largest open interval where the function f(x)=2x-5/x+3 is increasing AND concave up (at the same time)

Homework Equations

The Attempt at a Solution

I've run through the problem and arrived at a second derivative of f''(x)= -22/(x+3)^3. Setting this to zero I'm coming back with -3. However, I keep coming back with a situation in which this is concave down. Am I making some sort of error here?

 

[Solitare]

Pondering it, could it be as simple as this problem doesn't have a point of inflection; given that (-3+3)^2 returns as zero?

 

[ehild]

You have to find the interval where the function is concave up. x=-3 is a point where the function is neither concave up nor concave down, as you said it is not even defined. What does concave up mean?

ehild

 

[Solitare]

That there's a change of the sign upon reaching the point of inflection. In the case of concave up, from negative to positive.

 

[ehild]

It is the definition of an inflection point, that concavity changes sign. What do you know about the second derivative if the function is concave up? What is the concavity of f(x)=x^2 at x=1, for example?

ehild

 

[Solitare]

Perhaps I'm not following, but the second derivative of the function that you just posted is 2. There's no way to set that equal to 0.

 

[ehild]

How does f(x)=x^2 look like when you plot it? It is not a straight line. It is curved. Up or down? What is its concavity?

See: http://www.mathwords.com/c/concave_up.htm

ehild

 

[Solitare]

It's curved upward, so it's concave upward.

 

[ehild]

When curved up, the second derivative is positive. Where is the second derivative of your function positive?

ehild

 

[Solitare]

I'm not sure how this relates to my original question. In my original question I'm lost on whether or not the the function even has a point of inflection since my returned value when I set the second derivative to 0 is x=-3, which we've established isn't defined. Why we're going down this tangent of questioning is lost on me.

 

[ehild]

The original question is in what open interval is the function concave up. Nothing is said about the inflection point. A function changes concavity at an inflection point, but the concavity depends on the sign of the second derivative. It is concave up if the second derivative is positive.

ehild

 

[Solitare]

Correct. However, deriving to the second derivative and then setting to zero returns x = -3. That'd be the point of inflection, in which case I'd test the numbers around it to determine concavity. But as we've established, that's not defined in the function. Doesn't that mean that there's no points of inflection and thus no concavity?

 

[ehild]

Why do you set the second derivative to zero? ehild

 

[Solitare]

To find the value to test for concavity.

 

[ehild]

You find the point where the function is not defined. What about the concavity at x=-4? Does the function curved up or down there?

ehild

 

[Solitare]

At that point it doesn't, it's fairly constant right there.

 

[ehild]

Well, it is curved a bit, see again. And what about -3.5?

ehild

 

[Ray Vickson]

Homework Statement

1. Determine the largest open interval where the function f(x)=2x-5/x+3 is increasing AND concave up (at the same time)

Homework Equations

The Attempt at a Solution

I've run through the problem and arrived at a second derivative of f''(x)= -22/(x+3)^3. Setting this to zero I'm coming back with -3. However, I keep coming back with a situation in which this is concave down. Am I making some sort of error here?

As you wrote it, the function f(x) = 2x - 5/x + 3 has second derivative f''(x) = -10/x^3. Possibly you did NOT mean what you wrote; possibly you should have used brackets, to write f(x) = (2x-5)/(x+3). Is that what you meant? Certainly, that last f has the correct second derivative.

RGV

 

[Solitare]

Sorry, Ray. Yes. I meant (2x-5)/(x+3).

ehilde, if it's curving at all it's curving upward.

 

[ehild]

So it is concave up at x=-3.5, isn't it? And it changes concavity only at x=-3.
I can not help more. If you still do not get it, ask your teacher to explain concavity to you. ehild

 

[Solitare]

Ehild, you've not helped me one bit since we began speaking. Perhaps Ray would be more helpful at this point. I've made my question clear. I've performed the math as I've been taught. I've set the derivative equal to zero, which as we've said countless times now equals -3. I've selected random numbers both greater and lesser than -3. I've plugged those into the second derivative, as taught, and I've recorded the signs. Negative to the left of -3, positive to the right of -3. That, sir, is concave down if I've ever seen it. Again, all how I've been taught.

If -3 in the denominator makes it equal zero, then is not in the domain of f? I thought I was fairly clear in what I was asking.

 

[Ray Vickson]

Ehild, you've not helped me one bit since we began speaking. Perhaps Ray would be more helpful at this point. I've made my question clear. I've performed the math as I've been taught. I've set the derivative equal to zero, which as we've said countless times now equals -3. I've selected random numbers both greater and lesser than -3. I've plugged those into the second derivative, as taught, and I've recorded the signs. Negative to the left of -3, positive to the right of -3. That, sir, is concave down if I've ever seen it. Again, all how I've been taught.

If -3 in the denominator makes it equal zero, then is not in the domain of f? I thought I was fairly clear in what I was asking.

In {x < -3}, f(x) = (2x-5)/(x+3) has f''(x) > 0, so f is certainly not concave in that region. For {x > -3} we have f''(x) < 0, so f is concave. In both regions, f is strictly increasing; just plot it and see.

Personally, I don't like the terminology "concavity down or convex up", or whatever; I follow the practice in Operations Research and modern optimization, which just says "concave" and "convex". A concave function looks like (part of) a Macdonald's arch, while a convex function looks like (part of a) crater or a valley.

RGV

 

[ehild]

Personally, I don't like the terminology "concavity down or convex up", or whatever; I follow the practice in Operations Research and modern optimization, which just says "concave" and "convex". A concave function looks like (part of) a Macdonald's arch, while a convex function looks like (part of a) crater or a valley.

RGV

Hi Ray,

Neither like I the terminology "concave up", but that was the question of the original post. Convex is called sometimes "concave up"
http://www.mathwords.com/c/concave_up.htm

ehild

 

[Ray Vickson]

Hi Ray,

Neither like I the terminology "concave up", but that was the question of the original post. Convex is called sometimes "concave up"
http://www.mathwords.com/c/concave_up.htm

ehild

I know that was the terminology of the original problem, and that is why I explained what I meant by convex and concave (Macdonald arches, valleys, etc.) The OP can translate it into whatever terms he/she wants.

RGV

 

[ehild]

Solitare, I am sorry that I could not help.

ehild

 

[SithsNGiggles]

$f(x) = \frac{2x - 5}{x + 3}$
The graph obviously has a vertical asymptote x = -3.
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$f'(x) = \frac{11}{(x+3)^2}$

Find the critical points, which occur at values of x that make f '(x) zero or undefined.

Set $f'(x) = 0$:
$\frac{11}{(x+3)^2} \not= 0$
(so there are no points on the graph of f(x) for which the slope of the tangent line is zero)

Set the denominator equal to 0 to find what x makes f '(x) undefined:
$(x+3)^2 = 0$
$x = -3$

Apply the first derivative test.
On the interval $(-\infty, -3), f'(x) > 0$, which implies that f(x) is increasing over this interval.
On the interval $(-3, \infty), f'(x) > 0$, so f(x) is still increasing over this interval.

It should make sense to you that f is constantly increasing for the intervals on which f is defined, since the function has a positive integer in the numerator and a squared, positive quantity in the denominator.

- - - - - - - - - - - - - - - - - - - - - - - - - -

$f''(x) = - \frac{22}{(x+3)^3}$

Find x values for possible inflection points, which will occur for the x values that make f "(x) zero or undefined.

As with f '(x), f "(x) is never equal to zero.

Find the values that make f "(x) undefined:
As before, it's x = -3.

Set up some intervals and apply the second derivative test.
On the interval $(-\infty, -3), f''(x) > 0$. This is because any value within this interval, such as x = -4, gives you -22/(-negative number) = positive number. In the case of x = -4, f(-4) = (-22)/(-1)³ = 22. So, since f "(x) is positive, this means that f(x) is concave up over this interval.
On the interval $(-3, \infty), f''(x) < 0$, so f(x) is concave down over this interval.

And there you have it, f(x) is increasing over $(-\infty, -3) \cup (-3, \infty)$, but is also concave up only over $(-\infty, -3)$.

***NOTE: I'd like to add that the list of possible points of inflection are those x-values for which the concavity of f(x) may change, and not that f(x) is continuous at those points.***

 

[ehild]

Find x values for possible inflection points, which will occur for the x values that make f "(x) zero or undefined.

***NOTE: I'd like to add that the list of possible points of inflection are those x-values for which the concavity of f(x) may change, and not that f(x) is continuous at those points.***

You do not need inflection points to decide about the concavity of a function. For example, f(x)=x^2 is "concave up" (convex) everywhere, without having any inflection points.

ehild