Positive/Negative Intervals

[JSmith]

Determine the intervals on which the function f(x) = (0.25x - 4)(5x + 25)(x - 12) has:
a) a positive rate of change
b) a negative rate of change
c) neither

Are the intervals the zeros? for example, between the zeros, the graph is a parabola, therefore it has a positive rate of change up to the vertex and negative rate of change back to the next zero. This gives the interval between the two zeros neither a positive or negative rate of change. However, if you further break down the intervals further, you can have a positive and negative rate of change in their respective interval

 

[srmichael]

Determine the intervals on which the function f(x) = (0.25x - 4)(5x + 25)(x - 12) has:
a) a positive rate of change
b) a negative rate of change
c) neither

Are the intervals the zeros? for example, between the zeros, the graph is a parabola, therefore it has a positive rate of change up to the vertex and negative rate of change back to the next zero. This gives the interval between the two zeros neither a positive or negative rate of change. However, if you further break down the intervals further, you can have a positive and negative rate of change in their respective interval

Please show us the work you have done so far so that we know where you are stuck and therefore, where we can help you.

 

[JSmith]

I graphed the chart using the zeros, I just need to know what the question wants as far as the intervals. When we have talked about positive or negative intervals, we have just used the zeros to determine the intervals. However, this question is the rate of change, so I don't know if I should still use the zeros as intervals, or base my intervals on the positive or negative rates of change throughout the graph

 

[mmm4444bot]

Are [these] intervals [determined by] the zeros?

No -- because function f's zeros divide the domain into intervals where f(x) itself is either positive or negative.

This exercise is talking about the intervals where the slope of function f is positive or negative.

 

[mmm4444bot]

I just need to know what the question wants as far as the intervals.

It wants you to state the intervals where function f's rate is positive. (There are two such intervals.)

It wants you to state the interval where function f's rate is negative. (There is one such interval.)

It wants you to state where function f's rate is zero. (There are two points where this happens.)

However, this question is [about] the rate of change, so I don't know if I should
...
base my intervals on the positive or negative rates of change throughout the graph.

Yes -- in this exercise, it's the slope being positive, negative, or zero that matters, so consider the slope of function f.

That is, begin by looking at the curve of f's graph.

The values of f(x) do not matter. The values of the changing slope along the graph of f(x) matter.

Cheers :cool:

 

[HallsofIvy]

IF the problem is simply "on what intervals is the function value positive or negative" then the simplest thing to do is factor the coefficient of x out of each factor: f(x)= .25(x- 16)(5)(x+ 5)(x- 12)= (5/4)(x+ 5)(x- 12)(x- 16)= (5/4)(x-(-5))(x- 12)(x- 16).

The important facts here are:
(i) x- a is positive if x> a, negative if x< a
(ii) the product of an even number of negative factors is positive and the product of an odd number of negative factors is negative.

I we start with x< -5, then x is less than all of -4, 12, and 16 so all three factors are negative and f is negative. Then if -5< x< 12, x is larger than -5 so x- (-5)= x+ 5 is positive but x is still less than 12 and 16 so the last two factors are negative. Two negative factors means f is positive. If 12< x< 16, x is larger than both -5 and 12 so x+ 5 and x- 12 are positive but x< 16 so x- 16 is still negative. One negative factor means f is negative. Finally, if x> 16 all three factors are positive. There are no negative factors so f is positive.

However, you wrote "Determine the intervals on which the function f(x) = (0.25x - 4)(5x + 25)(x - 12) has:
a) a positive rate of change
b) a negative rate of change
c) neither"

You are NOT asked to determine where f itself is positive or negative but where it "rate of change" (derivative) is 0. This is a much harder problem and requires calculus, not algebra.

 

[JSmith]

It wants you to state the intervals where function f's rate is positive. (There are two such intervals.)

It wants you to state the interval where function f's rate is negative. (There is one such interval.)

It wants you to state where function f's rate is zero. (There are two points where this happens.)




Yes -- in this exercise, it's the slope being positive, negative, or zero that matters, so work with the first derivative of function f.

That is, begin by determining f`(x).

The values of f(x) do not matter. The values of f`(x) matter because f`(x) is the rate of f(x).

Cheers :cool:

I have the graph on my calculator and can see the two intervals where the graph is increasing, two where it has no rate of change, and the one where it is negative. How do I determine the exact x values for these intervals??

 

[JSmith]

Oops -- my bad. I was thinking that we were on the calculus board!

Does your class ever use graphing calculator's built-in features to determine specific x-values (eg: zooming, maximizing, solving) ?

Haha thanks I was slightly confused there! Yes, we do.

 

[mmm4444bot]

Oops -- my bad. I was thinking that we were on the calculus board! I edited-out my earlier references to f`(x).

Has your class ever used a graphing calculator's built-in features to determine specific x-values (eg: zooming, maximizing, minimizing, solving) ?

 

[mmm4444bot]

Haha thanks I was slightly confused there! Yes, we do.

Hey -- you can foresee the future!! (You responded before my post posted, heh, heh.)

Okay -- use technology to locate those x-values where the slope is zero (i.e., the top of the "hump" and the bottom of the "valley"). Those two x-values divide the Real number line into the intervals you seek.

You'll need to give decimal approximations for those x-values, as I believe that they are Irrational numbers. I generally round-off to 4 decimal places, if not specifically instructed otherwise.

Post your intervals, if you would like us to check your machine results. Cheers :cool:

 

[JSmith]

Hey -- you can foresee the future!! (You responded before my post posted, heh, heh.)

Okay -- use technology to locate those x-values where the slope is zero (i.e., the top of the "hump" and the bottom of the "valley"). Those two x-values divide the Real number line into the intervals you seek.

You'll need to give decimal approximations for those x-values, as I believe that they are Irrational numbers. I generally round-off to 4 decimal places, if not specifically instructed otherwise.

Post your intervals, if you would like us to check your machine results. Cheers :cool:

Ok! So the intervals I have are x<1.2289, 1.2289<x<14.104, x>14.104... I found the top of the "hump" to be at 1.2289. and the bottom of the "valley" to be 14.104. Did you find something similar?

 

[mmm4444bot]

So the intervals I have are

x < 1.2289

1.2289 < x < 14.1044

x > 14.1044

I found the top of the "hump" to be at x = 1.2289

and the bottom of the "valley" to be at x = 14.1044

Did you find something similar?

Yes, I found exactly the same! :cool:

 

[mmm4444bot]

The following is something that you would see, if you were to begin an introductory calculus course.

You know about slope (rate) of linear functions, already, yes? Lines have slope.

But curves also have slope! The slope generally changes at each and every point on a continuous curve.

When we talk about the rate of such "curved" function behavior, we're talking about these changing slopes.

And, when we talk about the specific slope of a curve, we're talking about the slope of a line tangent to the curve at a specific value of x.

Go to this page, and click on the yellow [next] button at the lower-right corner. Hopefully, you will see an animated tangent line showing the slope of the curve while advancing through increasing x-values.

If the animation does not work for you, let us know, and we will search for a better page.

Cheers :cool: