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**AlbertMThomas****Member**- Registered: 2013-08-21
- Posts: 9

Consider the function f(x)= 2x^4 - 512x² - 7x - 28. The function is plotted below. The tangents to the curve at the points A and C are shown. Find the value of the tangent at the point B if possible.

*Last edited by AlbertMThomas (2013-08-21 22:09:25)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,051

Hi;

Welcome to the forum.

The tangents to the curve at the points A and C are shown.

Where are they shown? If you want a numeric answer, I will need more.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,268

hi,

Judging by the other tangents, it looks like B is a point of inflexion. At such a point the dy/dx graph has a local minimum ( or maximum) and as a result the tangent there crosses the curve.

So get dy/dx and differentiate again and set to zero.

http://en.wikipedia.org/wiki/Inflection_point

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,051

Hi;

By golly, I think BB is right and I do not mean Big Brother. That does look like an inflection point at

see the red dot. Notice the change in the second derivative at the inflection point.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**AlbertMThomas****Member**- Registered: 2013-08-21
- Posts: 9

**A QUESTION: CAN A GRAPH contain MORE than one INFLECTION point? and WHY?**

**Here is the graph for the function:**

**Where is the inflection point in it??and If it doesn't contain infliction point how can I make sure of that as I don't see the whole graph??**

**What I tried, but I am sure this is a wrong answer:f(x)= 2x^4 - 512x² - 7x - 28f'(x)= 8x³ - 1024x - 7f''(x)= 24x² - 1024at inflection point f''(x)= 0 ⇒ 24x² - 1024 = 0x=±(8√6)/3then the value of the tangent at B is [f'(x)]"at x= (8√6)/3" = 8[(8√6)/3]³ - 1024[(8√6)/3] - 7 = -4466.159994**

**THIS IS WRONG ANSWER as it seems to be there is no inflection point**

*Last edited by AlbertMThomas (2013-08-22 00:19:49)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,051

I think that is because that is an undulation point, therefore it is not an inflection point.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**AlbertMThomas****Member**- Registered: 2013-08-21
- Posts: 9

Can we say it is impossible to find the slope, and why??

If this is an undulation then all what we did is wrong.. How can the solution be then??

The multiple choices for the answer are:

{A} 16

{B} 1

{C} 7

{D} 67

{E} It is impossible to say without knowing the exact value of the given information

What is the final answer???

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,051

Hi;

If those are the choices then I would go with E.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,268

There are points of inflexion at x = +/- 6.53...

These are the points where the gradient decreases to a minimum, and then increases once more.

My graph below for the function and its first derivative show this.

And I've labelled it B. But I'm assuming that was what the question is after.

Justification:

At A the tangent is above the line; at C it is below. So it's reasonable to assume that between A and C it makes the change.

Strictly it doesn't say that B is that point but what was the point of the question if the answer is just "We don't know" ?

So Ok, choose answer E, but under protest.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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