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Curve sketching

  • Thread starter Kolika28
  • Start date
86
12
Homework Statement
I'm given the function
f(x)=##2x^3+x^2-6x-2+sin(x)##
Homework Equations
MVT or IVT?
1.Prove that f'(x) is strictly decreasing at (- ##\infty##,a) and strictly increasing at (a,##\infty##).
2.Prove that f'(x) has exactly two roots.

I tried to find f''(x)=0, but I'm not able to solve the equation. What should I do?
 

PeroK

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Following the PF rules, let's see what you can do.
 
86
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Following the PF rules, let's see what you can do.
The thing is, I don't know how to use them (the PF rules) in this case. Yes, I can use IVT and plot in some values and find out that I have about two roots, but that does not prove that there is exactly two roots. When it comes to showing f'(x) is increasing and decreasing in the given intervals, I don't understand how I'am supposed to use MVT. I thought I had to solve the equation above for that( f''(x)=0 ), but I find it hard to solve.
 
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Homework Statement: I'm given the function
f(x)=##2x^3+x^2-6x-2+sin(x)##
Homework Equations: MVT or IVT?

1.Prove that f'(x) is strictly decreasing at (- ∞,a) and strictly increasing at (a,∞).
2.Prove that f'(x) has exactly two roots.
You could start by showing us what you got for f'(x). This will be a quadratic plus a cosine term. That cosine term always contributes something between -1 and +1. Investigate the quadratic part to see what it looks like if the cosine term isn't included, then look at the derivative f'(x) when the cosine term is included.
 

WWGD

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Kolika,
Compute the derivatives and use your knowledge about the relation between f' being 0, <0 or >0 in relation to whether the function is growing, etc. Mark the intervals where f' takes each value to start.
 

PeroK

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The thing is, I don't know how to use them (the PF rules) in this case. Yes, I can use IVT and plot in some values and find out that I have about two roots, but that does not prove that there is exactly two roots. When it comes to showing f'(x) is increasing and decreasing in the given intervals, I don't understand how I'am supposed to use MVT. I thought I had to solve the equation above for that( f''(x)=0 ), but I find it hard to solve.
##f''(x)=0## will likely be a transcendental equation and have no elementary solution. You can however sketch the graph to see how many roots it has and its sign.

You could use a numerical method to get an approximation for ##a##.
 
86
12
You could start by showing us what you got for f'(x). This will be a quadratic plus a cosine term. That cosine term always contributes something between -1 and +1. Investigate the quadratic part to see what it looks like if the cosine term isn't included, then look at the derivative f'(x) when the cosine term is included.
Sorry, I should have responend faster to everyones answers, but I have been really sick these past two days. So
##f'(x)=6x^2+2x-6+cosx##
You said I should investigate the quadratic part first:
##6x^2+2x-6=2(3x^2+x-3)##
##2(3x^2+x-3)=0##
##x=-1.18## or ##x=0.85##
I don't know how to look at the derivative when ##cos(x)## is included. What happens then?
##f''(x)=0## will likely be a transcendental equation and have no elementary solution. You can however sketch the graph to see how many roots it has and its sign.

You could use a numerical method to get an approximation for ##a##.
Kolika,
Compute the derivatives and use your knowledge about the relation between f' being 0, <0 or >0 in relation to whether the function is growing, etc. Mark the intervals where f' takes each value to start.
My teacher said that he didn't want us to sketch the graph, because he means we should argument with theorms and logic.
 
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So ##f'(x)=6x^2+2x-6+cosx##
The graph of f' will look somewhat like the graph of y = 6x^2 + 2x - 6, but with some oscillation due to the cosine term. The graph of y = 6x^2 + 2x -6 is a parabola, right? Which way does it open, up or down? Where is its maximum or minimum point?
The graph of f' can't vary by more than +1 or -1 from the graph of the parabola, so maybe you can get at least a rough idea of where f' is positive and where it is negative.
My teacher said that he didn't want us to sketch the graph, because he means we should argument with theorms and logic.
What I'm suggesting is based on logic, as well as possibly one or the other of the theorems you cited in post #1.
 

OmCheeto

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My teacher said that he didn't want us to sketch the graph, because he means we should argument with theorms and logic.
Perhaps someone should change the thread title to "Not Curve Sketching".
The first thing I did to solve this problem was sketch the curve{s}.
 
86
12
The graph of f' will look somewhat like the graph of y = 6x^2 + 2x - 6, but with some oscillation due to the cosine term. The graph of y = 6x^2 + 2x -6 is a parabola, right? Which way does it open, up or down? Where is its maximum or minimum point?
The graph of f' can't vary by more than +1 or -1 from the graph of the parabola, so maybe you can get at least a rough idea of where f' is positive and where it is negative.
What I'm suggesting is based on logic, as well as possibly one or the other of the theorems you cited in post #1.
But if I do so, I find exact values for x, and my teacher said that it is not necessary. I should only prove that there exists exactly three solutions.
 
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My teacher said that he didn't want us to sketch the graph
I'm not advising you to sketch the graph of ##f(x) = 2x^3 + x^2 - 6x - 2 + \sin(x)##. I'm advising you to sketch a graph of the derivative function. I don't believe your teacher said not to do that.

But if I do so, I find exact values for x, and my teacher said that it is not necessary. I should only prove that there exists exactly three solutions.
From the graph of the derivative, f', you can get a good idea of where the graph of f is increasing or decreasing, and from that, you can get some idea about where the x-intercepts of f are, and how many there are.
 

epenguin

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As helpers here are not using your exact same textbook not all of them will know what you mean by IVT or MVT. OK I am guessing intermediate value theorem and maximum value theorem. Not a uite sure what those are though.

You talk about solving an equation. Actually all you have to do is establish an inequality.
There are certainly maximum and minimum values of sines and cosines that you might remember.
 
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not all of them will know what you mean by IVT or MVT. OK I am guessing intermediate value theorem and maximum value theorem.
Yes and no. MVT is almost certainly mean value theorem.
There are certainly maximum and minimum values of sines and cosines that you might remember.
Alluded to in posts 4 and 8.
 

scottdave

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Sorry, I should have responend faster to everyones answers, but I have been really sick these past two days. So
##f'(x)=6x^2+2x-6+cosx##
You said I should investigate the quadratic part first:
##6x^2+2x-6=2(3x^2+x-3)##
##2(3x^2+x-3)=0##
##x=-1.18## or ##x=0.85##
I don't know how to look at the derivative when ##cos(x)## is included. What happens then?


My teacher said that he didn't want us to sketch the graph, because he means we should argument with theorms and logic.
I'm just coming in. I haven't read all the posts yet. There are some good ideas here. So if you don't sketch it what can you do? How about consider the cubic with the sine term removed. See if you can find local minima or Maxima. Then look to see if a sine term (which must be between +1 and -1) would cause to cross the axis. Try different things and see what results you get. Is what you tried something which could be used in a proof?
 

vela

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First, I would recast the question to be in terms of ##g(x) = f'(x)##. Show that g(x) is strictly increasing..., etc. The fact that g is the derivative of f is irrelevant, but it can be confusing.

Try proving the claims for ##h(x)=6x^2+2x-6##. That should be pretty straightforward. If you don't see how to do that, that's where you should start. (Hint: you don't need to find the roots of the ##h(x)## like you did earlier.)

Once you have that down, then consider the effect of adding the cosine term. How will that affect the logic of your previous proofs?
 

epenguin

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My teacher said that he didn't want us to sketch the graph, because he means we should argument with theorms and logic.
That needn't stop you sketching it anyway, it could lead to some insight or understanding, after which you could argue with theorems and logic only.
 
86
12
Thank you for everyones help. It took two weeks for me to understand it, but it's better late than never!
 

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