Intervals

[epicjacob1123]

I have a function f (x) I know that:

f(x) = 1 + a/x + a/x^2

f'(x) = -((x+2)*a)/x^3

f''(x) = ((2x +6)*a)/x^4

Where a>0. I have to find all intervals where the function is increasing, decreasing, concave up and concave down. I understand how to identify places in a function where it is increasing/ decreasing and concave up/down, but not ALL intervals.

 

[Deleted member 4993]

I have a function f (x) I know that:

f(x) = 1 + a/x + a/x^2

f'(x) = -((x+2)*a)/x^3

f''(x) = ((2x +6)*a)/x^4

Where a>0. I have to find all intervals where the function is increasing, decreasing, concave up and concave down. I understand how to identify places in a function where it is increasing/ decreasing and concave up/down, but not ALL intervals.

Start with plotting:

g(x) = 1/x + 1/x²

What are the critical points of g(x)?

At which points this function is increasing, decreasing, concave up and concave down?

How is f(x) = 1 + a/x + a/x^2 different from g(x)?

 

[stapel]

I understand [stuff] but not ALL intervals.

What do you mean when you say that you don't understand "ALL intervals"? Are you having trouble figuring out what the intervals (of increase / decrease and concavity) are?

Thank you!

 

[epicjacob1123]

What do you mean when you say that you don't understand "ALL intervals"? Are you having trouble figuring out what the intervals (of increase / decrease and concavity) are?

Thank you!

Yes, I don't understand how to find the intervals where the graph is increasing/ decreasing when I don't know what the value a is, other than it is >0.

 

[stapel]

Yes, I don't understand how to find the intervals where the graph is increasing/ decreasing when I don't know what the value a is, other than it is >0.

Well, we do know that "a" is positive. This let's us know that we don't have to worry about the value of, say, "ax" depending on anything other than the sign on the term (the "plus" or "minus" in front of it). If we're looking at, say, x = 2, then we know that "ax" must be positive.

You have the following:

. . . . .\(\displaystyle f(x)\, =\, \dfrac{x^2\, +\, ax\, +\, a}{x^2}\)

. . . . .\(\displaystyle f'(x)\, =\, \dfrac{-a(x\, +\, 2)}{x^3}\)

. . . . .\(\displaystyle f"(x)\, =\, \dfrac{2a(x\, +\, 3)}{x^4}\)

(I haven't checked your work on these. I'm assuming you did the math correctly.)

To find, for instance, the intervals of sign for concavity, do the usual thing: Set the second derivative equal to zero (or, which is the same thing in this case, set the numerator equal to zero), and solve:

. . . . .\(\displaystyle 2a(x\, +\, 3)\, =\, 0\)

. . . . .\(\displaystyle x\, =\, -3\)

As you can see in this case, the value of "a" is irrelevant. Now, note that you can't have x equal to zero, and set the intervals:

. . . . .\(\displaystyle (-\infty,\, -3),\, (-3,\, 0),\, (0,\, +\infty)\)

Now look at the factors, just as you did back in algebra when you were working with solving rational inequalities (refresher):

You know that, in all cases (other than x = 0), x⁴ must be positive. You know that 2a is, in all cases, positive. So you only care about the "x + 3" factor. Where x < -3, then x + 3 < 0, so the sign on the whole expression is... what? Everywhere else, the sign on the whole expression is... what? So the concavity on each of the three intervals must be... what?

Now do the same thing for f(x) and f'(x).

 

[epicjacob1123]

Well, we do know that "a" is positive. This let's us know that we don't have to worry about the value of, say, "ax" depending on anything other than the sign on the term (the "plus" or "minus" in front of it). If we're looking at, say, x = 2, then we know that "ax" must be positive.

You have the following:

. . . . .\(\displaystyle f(x)\, =\, \dfrac{x^2\, +\, ax\, +\, a}{x^2}\)

. . . . .\(\displaystyle f'(x)\, =\, \dfrac{-a(x\, +\, 2)}{x^3}\)

. . . . .\(\displaystyle f"(x)\, =\, \dfrac{2a(x\, +\, 3)}{x^4}\)

(I haven't checked your work on these. I'm assuming you did the math correctly.)

To find, for instance, the intervals of sign for concavity, do the usual thing: Set the second derivative equal to zero (or, which is the same thing in this case, set the numerator equal to zero), and solve:

. . . . .\(\displaystyle 2a(x\, +\, 3)\, =\, 0\)

. . . . .\(\displaystyle x\, =\, -3\)

As you can see in this case, the value of "a" is irrelevant. Now, note that you can't have x equal to zero, and set the intervals:

. . . . .\(\displaystyle (-\infty,\, -3),\, (-3,\, 0),\, (0,\, +\infty)\)

Now look at the factors, just as you did back in algebra when you were working with solving rational inequalities (refresher):

You know that, in all cases (other than x = 0), x⁴ must be positive. You know that 2a is, in all cases, positive. So you only care about the "x + 3" factor. Where x < -3, then x + 3 < 0, so the sign on the whole expression is... what? Everywhere else, the sign on the whole expression is... what? So the concavity on each of the three intervals must be... what?

Now do the same thing for f(x) and f'(x).

Ohhh I see. Thanks for the help. It's much easier than I thought.