Turning Points

mathdad

mathdad

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Find the turning points of f(x) = x^2(x - 2).

I set each factor to 0 and solve for x.

x^2 = 0

sqrt{x^2} = sqrt{0}

x = 0

x - 2 = 0

x = 2

Turning points are x = 0 and x = 2.

Yes?
 
mathdad said:
Find the turning points of f(x) = x^2(x - 2)

I set each factor to 0 and solve for x ...
Click to expand...
No. That's how we find the x-intercepts (0 and 2).

Turning points are covered in calculus. From where does this exercise come?

PS: Did you look at a graph of the function? That's a quick way for you to see that something's not correct. Google will plot it for you; enter the following.

plot y=x^2*(x-2)

?
 
Otis said:
No. That's how we find the x-intercepts (0 and 2).

Turning points are covered in calculus. From where does this exercise come?

PS: Did you look at a graph of the function? That's a quick way for you to see that something's not correct. Google will plot it for you; enter the following.

plot y=x^2*(x-2)

?
Click to expand...

I will check youtube.com for clips. I was told to find the zeros. Afterward, plot the zeros on the number line. Select a number from different intervals on the divided number line to evaluate the given function.

Negative answers means the graph is decreasing and positive answers means the function is increasing. I was also told to use the multiplicity of the function to determine if the graph crosses or touches the x-axis.
 
mathdad said:
I will check youtube.com for clips. I was told to find the zeros. The zeros (or roots) of and equation are the values of x that make the function 0, ie y=0. Graphically, the zeros or roots are the x-intercepts (because that is where y=0.

Afterward, plot the zeros on the number line. Select a number from different intervals on the divided number line to evaluate the given function.

Negative answers means the graph is decreasing No. Negative answers mean the graph is below the x-axis and positive answers means the function is increasing. No Positive answers mean the graph is above the x-axis.

I was also told to use the multiplicity of the function to determine if the graph crosses or touches the x-axis.
You probably mean multiplicity of the factors. Do you know what that means?
Click to expand...
What was the exact wording of this question? Are you after a quick sketch of the graph?
 
Harry_the_cat said:
What was the exact wording of this question? Are you after a quick sketch of the graph?
Click to expand...

The actual wording of the question is back home in my textbook. I will post the question in its entirety later this morning after work. This question has 5 parts (a through e).
 
Otis said:
No. That's how we find the x-intercepts (0 and 2).

Turning points are covered in calculus. From where does this exercise come?

PS: Did you look at a graph of the function? That's a quick way for you to see that something's not correct. Google will plot it for you; enter the following.

plot y=x^2*(x-2)

?
Click to expand...

This is what I am looking for:

 
What? "End behavior" has nothing to do with "turning points". A function has a "turning point" where the graph either turns from going up to going down or turns from going down to going up. As long as the graph is going up, the derivative is positive. As long as the graph is going down the derivative is negative. At a turning point the derivative changes sign. Since a derivative always satisfies the "intermediate value property", at a turning point the derivative must be zero, not the function itself.

Caution: the derivative being 0 is not enough- the function \(\displaystyle y= x^3\) has derivative \(\displaystyle y'= 3x^2\) has derivative 0 at x= 0 but the derivative does not change sign there!
 
HallsofIvy said:
What? "End behavior" has nothing to do with "turning points". A function has a "turning point" where the graph either turns from going up to going down or turns from going down to going up. As long as the graph is going up, the derivative is positive. As long as the graph is going down the derivative is negative. At a turning point the derivative changes sign. Since a derivative always satisfies the "intermediate value property", at a turning point the derivative must be zero, not the function itself.

Caution: the derivative being 0 is not enough- the function \(\displaystyle y= x^3\) has derivative \(\displaystyle y'= 3x^2\) has derivative 0 at x= 0 but the derivative does not change sign there!
Click to expand...

1. I am reviewing college algebra not calculus.

2. Look at the posted video clip. This is exactly what Sullivan talks about in his textbook.
 
mathdad said:
Find the turning points of f(x) = x^2(x - 2).

I set each factor to 0 and solve for x.

x^2 = 0

sqrt{x^2} = sqrt{0}

x = 0

x - 2 = 0

x = 2

Turning points are x = 0 and x = 2.

Yes?
Click to expand...
This will repeat points previously made, but I want to put them together in reference to your original post.

Finding turning points of differentiable functions like polynomials is essentially a type of problem dealt with in differential calculus. There may be algebraic ways to find turning points of polynomials of degree > 2, but there is virtually no practical benefit to learning them. Did you describe the problem correctly? There is a reason that we ask that problems be quoted EXACTLY and completely. That reason is that students frequently paraphrase a problem in an incorrect way, which in turn prevents them from solving the problem. Did the problem say "turning points" or something else?

What you did was to find the zeroes of the function. If that is what the problem asked for, then you got the correct answer using a method that will work for SOME functions.

Finding the zeroes of a continuous function will also let you determine the INTERVAL OR INTERVALS in which turning points of that function will be found and determine whether each turning point is a local minimum or a local maximum. This information is sufficient to let you sketch a crude graph.

HOWEVER, finding an interval is not the same as finding a point. One answer looks like "the function has a local minimum somewhere between 10 and 32," and the other looks like "the function has a local minimum at 19." One answer is relatively vague; the other is not.

In a later post, you show a video about the end behavior of polynomials. This has nothing at all to do with finding turning points or the intervals in which turning points occur.

I have tried to stress that definitions are very important in math; they have been developed to clear up what were frequently years of confusion. If you do not understand a definition in a text, ask about it before frustrating yourself trying problems.
 
Last edited:
JeffM said:
This will repeat points previously made, but I want to put them together in reference to your original post.

Finding turning points of differentiable functions like polynomials is essentially a type of problem dealt with in differential calculus. There may be algebraic ways to find turning points of polynomials of degree > 2, but there is virtually no practical benefit to learning them. Did you describe the problem correctly? There is a reason that we ask that problems be quoted EXACTLY and completely. That reason is that students frequently paraphrase a problem in an incorrect way, which in turn prevents them from solving the problem. Did the problem say "turning points" or something else?

What you did was to find the zeroes of the function. If that is what the problem asked for, then you got the correct answer using a method that will work for SOME functions.

Finding the zeroes of a continuous function will also let you determine the INTERVAL OR INTERVALS in which turning points of that function will be found and determine whether each turning point is a local minimum or a local maximum. This information is sufficient to let you sketch a crude graph.

HOWEVER, finding an interval is not the same as finding a point. One answer looks like "the function has a local minimum somewhere between 10 and 32," and the other looks like "the function has a local minimum at 19." One answer is relatively vague; the other is not.

In a later post, you show a video about the end behavior of polynomials. This has nothing at all to do with finding turning points or the intervals in which turning points occur.

I have tried to stress that definitions are very important in math; they have been developed to clear up what were frequently years of confusion. If you do not understand a definition in a text, ask about it before frustrating yourself trying problems.
Click to expand...

You have made your message clear to me. However, I will use youtube.com video clips to help me with the rest of my study of college algebra. I will use this site mainly for help with word problems.
 
mathdad said:
The actual wording of the question is back home in my textbook. I will post the question in its entirety later this morning after work.
Click to expand...
I'd still like to see the exercise, mathdad. Did you forget to post it?

\(\;\)
 
Otis said:
I'd still like to see the exercise, mathdad. Did you forget to post it?

\(\;\)
Click to expand...

Sorry but I have been very busy.

Here is the actual question:

Given f(x) = (-2x^2)(x^2 - 2), solve A through E.

A. List each real zero and its multiplicity.
B. Determine if the graph crosses or touches the x-axis at each x-intercept.
C. Determine the behavior of the graph near each x-intercept (zero).
D. Determine the maximum number of turning points on the graph.
E. Determine the end behavior; that is, find the power function that the graph of f resembles for large values of | x |.

Sorry for posting the wrong question.
 
I would begin by writing the function as:

[MATH]f(x)=-2x^2(x+\sqrt{2})(x-\sqrt{2})[/MATH]
Now it's real easy to see each zero and its multiplicity. Can you state them?
 
MarkFL said:
I would begin by writing the function as:

[MATH]f(x)=-2x^2(x+\sqrt{2})(x-\sqrt{2})[/MATH]
Now it's real easy to see each zero and its multiplicity. Can you state them?
Click to expand...

Zeros: x = 0, -sqrt{x}, sqrt{x}

Multiplicity for 0 is 2.
Multiplicity for -sqrt{x} and sqrt{x} is 1.
 
Okay, good. Now, based on the parity of the multiplicities of the zeros, can you answer part B?
 
mathdad said:
Zeros: x = 0, -sqrt{x}, sqrt{x}

Multiplicity for 0 is 2.
Multiplicity for -sqrt{x} and sqrt{x} is 1.
Click to expand...
Just to make sure you (and readers) recognize what I take to be a typo, I think you meant this:

Zeros: x = 0, [MATH]-\sqrt{2}[/MATH], [MATH]\sqrt{2}[/MATH]​
​
Multiplicity for 0 is 2.​
Multiplicity for [MATH]-\sqrt{2}[/MATH] and [MATH]\sqrt{2}[/MATH] is 1.​
 
mathdad said:
... Determine the maximum number of turning points ...

Sorry for posting the wrong question
Click to expand...
This is a good example of why the forum guidelines ask people to post given information word-for-word. If you had followed the guidelines, we could have told you right away that you misread part D. You don't need to find the turning points.

\(\;\)
 
Dr.Peterson said:
Just to make sure you (and readers) recognize what I take to be a typo, I think you meant this:

Zeros: x = 0, [MATH]-\sqrt{2}[/MATH], [MATH]\sqrt{2}[/MATH]​
​
Multiplicity for 0 is 2.​
Multiplicity for [MATH]-\sqrt{2}[/MATH] and [MATH]\sqrt{2}[/MATH] is 1.​
Click to expand...

Is this not what I posted?
 
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