Horizontal pt. of inflex. (1 Viewer)

[BlueGas]

I need help for this question, the answers were saying that "to the left of D tangent +ve", but how when it's clearly a negative gradient? Anyway I also wanna know what's the difference between vertical and horizontal pt of inflex. and how would I do this question.

 

[davidgoes4wce]

I look at the stationary points A,B,C,D a good way to look at it (keeping it in terms of simple terminology), is look at the slopes to the left and right of the stationary points.

For example for A, the slope to the left of the curve is negative, and to the right of A is positive.

For example for B, the slope to the left of the curve is positive, and to the right of B is positive.

For example for C, the slope to the left of the curve is positive, and to the right of C is negative.

For example for D, the slope to the left of the curve is negative, and to the right of D is positive.

For a horizontal point of inflection to occur in this example, you have to watch for instances where the slope of the curves does not change going from left to right.

Hope I made things easier.

 

[BlueGas]

So for a horizontal pt of inflex to occur the slopes on both sides must be negative/negative or positive/positive?

 

[InteGrand]

So for a horizontal pt of inflex to occur the slopes on both sides must be negative/negative or positive/positive?

Yes (in other words, a point where the slope of the curve is 0, but it's not a local max./min.).

 

[BlueGas]

Yes (in other words, a point where the slope of the curve is 0, but it's not a local max./min.).

Okay so B is positive/positive, the answer should be B, but it isn't, the answer is D.

 

[rand_althor]

Okay so B is positive/positive, the answer should be B, but it isn't, the answer is D.

That's because this is a graph of the first derivative, not the function itself. The first derivative is positive on both sides of D.

 

[kawaiipotato]

^
Note that also, horizontal points occur when f'(x) = 0 AND f"(x) = 0
Which means its the stationary point for the f'(x) and also when it cuts the x axis for the f'(x) curve

 

[BlueGas]

That's because this is a graph of the first derivative, not the function itself. The first derivative is positive on both sides of D.

How is it positive on both side of D? The left side of D starts from top left and ends up at bottom, and that's a negative, isn't it?

 

[rand_althor]

How is it positive on both side of D? The left side of D starts from top left and ends up at bottom, and that's a negative, isn't it?

on both sides of D. You're looking at the gradient of the first derivative, not the gradient of the function.

 

[BlueGas]

So am I meant to draw the f(x) of this graph since the question is showing the f'(x)? Well how would I graph f(x)?

 

[rand_althor]

No you don't need to. The conditions for a horizontal point of inflexion are:

1. 
2. 
3. is the same on either side

Clearly f'(D)=0 as it is on the x-axis. Since D is also a stationary point, that means f''(D)=0. Now the y-axis in this graph is f'(x). f'(x) is greater than 0 on either side of D, i.e. it is positive on both sides of D.

 

[BlueGas]

No you don't need to. The conditions for a horizontal point of inflexion are:

1. 
2. 
3. is the same on either side

Clearly f'(D)=0 as it is on the x-axis. Since D is also a stationary point, that means f''(D)=0. Now the y-axis in this graph is f'(x). f'(x) is greater than 0 on either side of D, i.e. it is positive on both sides of D.

How about a vertical pt of inflex?

 

[InteGrand]

How about a vertical pt of inflex?

 

[BlueGas]

Could you try explaining it without the infinity and approaching stuff? That confuses me.

 

[InteGrand]

Could you try explaining it without the infinity and approaching stuff? That confuses me.

Basically, in order to have a vertical point of inflexion on a smooth curve, it is necessary and sufficient to find a point on the curve where the following hold:

- the slope is vertical (i.e. the f'(x) has to approach infinity; this is what we mean by vertical tangent)
- the concavity changes from concave up to concave down, or vice versa.

An example is x = 0 on f(x) = x^(1/3). This is because if we look at the graph of this function, we can see that at x = 0, the slope is vertical (to prove this, you can show that f'(x) approaches infinity or minus infinity as we approach x = 0), and the concavity changes about this point (the curve is concave up to the left of it, and concave down to the right of it).

(Here's a graph of the function: http://amsi.org.au/ESA_Senior_Years/imageSenior/2b_27.png)

 

[BlueGas]

No you don't need to. The conditions for a horizontal point of inflexion are:

1. 
2. 
3. is the same on either side

Clearly f'(D)=0 as it is on the x-axis. Since D is also a stationary point, that means f''(D)=0. Now the y-axis in this graph is f'(x). f'(x) is greater than 0 on either side of D, i.e. it is positive on both sides of D.

So why can't the answer be B? It's on the x-axis so f(x) = 0, and it's a stationary point so f''(x) = 0, or is it because there is a part of B that's under the x-axis?

 

[rand_althor]

So why can't the answer be B? It's on the x-axis so f(x) = 0, and it's a stationary point so f''(x) = 0, or is it because there is a part of B that's under the x-axis?

Since it is on the x-axis in this graph, f'(x)=0, not f(x). Also, just before B, f'(x)<0 so the gradient is negative, and right after B, f'(x)>0 so the gradient is positive. So f'(x) isn't the same on either side.

 

[InteGrand]

 

[InteGrand]