You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

**The graph of f(x) = ax^3 + bx^2 + cx + d has a relative maximum at (0,0) and a point of inflection at (1, -2). Find the equation of f by determining a,b,c and d.**

This short sentance is absolutely loaded with clues.

First lets find the first and second derivative of f(x)

f(x) = ax^3 + bx^2 + cx + d

f'(x) = 3ax^2 + 2bx + c

f''(x) = 6ax + 2b

Now the problem stated the function has a relative maximum at (0,0), this tells us two things at once. First it tells us that f(0) = 0 and also that f'(x) = 0

f(0) = a(0^3) + b(0^2) + 0c + d = 0 so d must equal 0.

f'(0) = 3a(0^2) + 2b(0) + c = 0 so c must also equal

We are then told it has an inflection point at (1,-2), this means f(1) = -2 and if I remember correctly, the second derivative equals 0 at an inflection point, so f''(1) = 0

f(1) = a + b + c + d = -2 we know c and d are zero so: a + b = -2

f''(1) = 6a + 2b = 0 so -3a = b

-3a = b and a + b = -2

a = 1

b = -3

so f(x) = x^3 - 3x^2

Cool problem huh?

Now down to the main question, the problem stated there was an inflection point at (1,-2) this means the second derivative will be 0 at 1, but doesn't this also mean the first derivative will be 0 as well?

Take for instance the function f(x) = x^3

f'(x) = 3x^2

f''(x) = 6x

f''(0) is 0 since it is an inflection point, but also f'(0) = 0. When first trying to solve the problem, I thought the fact that f had an inflection point at (1,-2) also suggested that f'(1) = 0. But it isn't, and it screwed the problem up.

Thats wierd. I really would have thought an inflection point would have a derivative of 0 at that point. Or do I faintly remember reading that the derivative of a function does not exist at an inflection point?

Agh... so much to remember.. I'm forgeting what I read about inflection points. I think I remember reading that the second derivative of an inflection point is always 0, but the reverse is not always true. There is not always an inflection point for values of x for which the second derivative equals zero.

A logarithm is just a misspelled algorithm.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

You're right, points of inflection have f'(x) and f''(x) both equal to 0.

But if you draw the graph according to what you've found, you actually get a local maximum at (0,0) and a local minimum at (2,-4).

The reason is that the question is impossible. All graphs can only have a number of turning points equal to one less than their highest power. Points of inflection count as 2 turning points, so the question defines the graph as having 3 turning points, even though it can only have 2.

Try solving it for ax^4 + bx^3 + cx^2 + dx + e.

If you use f'(1) = 0, then you've got the fifth equation to deal with the fifth variable.

Why did the vector cross the road?

It wanted to be normal.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

You're right, points of inflection have f'(x) and f''(x) both equal to 0.

Huh? Inflection points don't have to have a slope of 0. On the contrary, they *can't* have a slope of 0.

Check this over, make sure I didn't make any mistakes:

f(x) = ax^3 + bx^2 + cx + d

f'(x) = 3ax^2 + 2bx + c

f''(x) = 6ax + 2b

We want an inflection point at (1, -2).

f''(1) = 6a + 2b = 0

3a + b = 0, or b = -3a

Now we have a max at (0,0):

f'(0) = c = 0

Finally, we want the graph to pass through the point (0, 0):

f(0) = d = 0

Since c is 0 in the 2nd derivative, c is 0 in the original function making it:

f(x) = ax^3 + bx^2

The only restricting factor on this equation is:

b = -3a

Trying it for a = 1:

b = -3, a = 1:

f(x) = x^3 - 3x^2

This passes through the points (0, 0) and (1, -2), has a local max at (0, 0), and an inflection at (1, -2).

*Last edited by Ricky (2006-01-11 04:05:08)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Here is a graph of x^3 and cos(x^3).

The slope is zero at the inflection point of x^3 at the origin, but the slope of the inflection points of the cos(x^3) are slanted slopes along the x-axis where the function crosses the x-axis.

So inflection points can have slopes other than zero, but as you said, the second derivative is zero

because that says the slope is not changing at that point.

The slope is in limbo for a second.

**igloo** **myrtilles** **fourmis**

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Oops. OK then. I always learnt that points of inflection were stationary points that weren't extrema. Just ignore that post up there.

Why did the vector cross the road?

It wanted to be normal.

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

What defines an inflection point? Generally I think its when the slope briefly decreases and the increases again, and maybe vice versa as well.

A logarithm is just a misspelled algorithm.

Offline

**God****Member**- Registered: 2005-08-25
- Posts: 59

Inflection point is when the slope changes from increasing to decreasing or vice versa via 0. In other words, f''(x) = 0 and f'''(x) is not equal to zero. Another way of putting it is that the function changes concavity.

Offline

**irspow****Member**- Registered: 2005-11-24
- Posts: 456

What defines an inflection point is that both the first and second derivatives equal zero. The velocity and the acceleration are zero at a turning point, specifically an inflection is the moment when the velocity or rate of change is reversing direction.

When working on the above question it is found that the function does not exist. Mikau correctly eliminated c and d in the beginning, but failed to use both derivatives together to eliminate a and b as well.

f(x) = ax³ + bx²

f'(x) = 3ax² + 2bx

f''(x) = 6ax + 2b

Because an inflection occurs at x=1;

3a + 2b = 0 and 6a + 2b = 0

3a + 2b = 6a + 2b, which only works if both a and b are zero.

y = 0 does not satisfy any of the conditions originally proposed.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Another way of putting it is that the function changes concavity.

Yes.

What defines an inflection point is that both the first and second derivatives equal zero.

It is not possible to change concavity when the slope is 0.

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

a point on a curve at which the second derivative changes sign. This is very similar to the previous definition, since the sign of the curvature is always the same as the sign of the second derivative, but note that the curvature is not the same as the second derivative.

Or:

a point (x,y) on a function, f(x), at which the first derivative, f'(x), is at an extremum, i.e. a minimum or maximum. (This is not the same as saying that y is at an extremum, and in fact implies that y is not at an extremum).

Or:

if f'(x) is not zero, the point is a non-stationary point of inflection

*Last edited by Ricky (2006-01-11 09:50:47)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**God****Member**- Registered: 2005-08-25
- Posts: 59

The first derivative doesn't have to be 0 to have a PoI.

For example, the function f(x) = x^3 - 3x has inflection point when x = 0 because the following two conditions are met when x = 0:

f''(x) = 6x, which is 0 when x = 0

f'''(x) = 6, is not equal to 0 when x = 0

Note that although f'(x) = 3x^2 - 3 and f'(0) = -3, the function still changes concavity.

Offline

**irspow****Member**- Registered: 2005-11-24
- Posts: 456

Ricky, thank you for correcting me. I was confusing the second derivative test for local extrema. (blush)

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

f'''(x) = 6, is not equal to 0 when x = 0

God, I have never heard of that. On wikipedia, it states:

One also needs the lowest-order non-zero derivative to be of odd order (third, fifth, etc.). If the lowest-order non-zero derivative is of even order, the point is not a point of inflection. (An example of such a function is y = x^4 -x).

It sounds similar, but not quite the same.

Edit:

But the wikipedia article doesn't sound right either. According to it:

f(x) = x^4 - 5x^2 + 4

Doesn't have any inflection points, yet it appears to change concavity.

*Last edited by Ricky (2006-01-11 10:23:59)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**God****Member**- Registered: 2005-08-25
- Posts: 59

Wikipedia article: Inflection point

a point on a curve at which the curvature changes sign. The curve changes from being concave upwards (positive curvature) to concave downwards (negative curvature), or vice versa. If one imagines driving a vehicle along the curve, it is a point at which the steering-wheel is momentarily 'straight', being turned from left to right or vice versa.

Offline

Pages: **1**