What is the maximum slope for this line?

In summary: I derived x2e-3x/xI simplified it to e-3x(2-3x-1) = 0.So my x-value is 1/3. But when I look at it on the graph, it looks like that value is a local minimum...so I'm confused.Great! That's the answer I got as well. Are you sure you aren't looking at a graph of e-3x(1-3x), instead of x2e-3x?
  • #1
zJakeAdam
2
0

Homework Statement



[PLAIN]http://img824.imageshack.us/img824/5387/idkz.jpg

Homework Equations





The Attempt at a Solution



I really don't even understand where to start. I mean, the graph just seems to go continuous and so wouldn't that mean their is no maximum slope?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Greetings and welcome!

zJakeAdam said:
I mean, the graph just seems to go continuous and so wouldn't that mean their is no maximum slope?

Interesting, are you sure you graphed it correctly? Consider the formula for the slope of a line between two points:

[tex] slope = \frac{y_2 - y_1}{x_2 - x_1} [/tex].

Since we know that the line in question goes through the origin, we can say (x1,y1) = (0,0). Now consider some point on the curve, (x2, y2) = (x, x2e-3x). This should make sense because when we are x units to the right, the curve is x2e-3x units up. Try plugging these values into the slope formula and then maximizing the slope.
 
  • #3
Undoubtedly0 said:
Greetings and welcome!



Interesting, are you sure you graphed it correctly? Consider the formula for the slope of a line between two points:

[tex] slope = \frac{y_2 - y_1}{x_2 - x_1} [/tex].

Since we know that the line in question goes through the origin, we can say (x1,y1) = (0,0). Now consider some point on the curve, (x2, y2) = (x, x2e-3x). This should make sense because when we are x units to the right, the curve is x2e-3x units up. Try plugging these values into the slope formula and then maximizing the slope.

Hmm...can you explain some of the steps?

I derived x2e-3x/x

I simplified it to e-3x(2-3x-1) = 0.

So my x-value is 1/3. But when I look at it on the graph, it looks like that value is a local minimum...so I'm confused.
 
  • #4
Great! That's the answer I got as well. Are you sure you aren't looking at a graph of e-3x(1-3x), instead of x2e-3x?
 

1. What is the maximum slope for a line?

The maximum slope for a line is infinite, which means the line is vertical. This is because the slope of a vertical line is undefined.

2. How is the maximum slope calculated?

The maximum slope is calculated by finding the slope of the steepest line possible on a given graph. This can be done by finding the slope of two points that are located at the highest and lowest points on the graph.

3. What is the significance of the maximum slope?

The maximum slope is significant because it represents the steepest incline or decline of a line. This can be important in various real-world applications, such as determining the maximum grade for a road or the maximum slope for a ski slope.

4. Can the maximum slope be negative?

No, the maximum slope cannot be negative. Slope is a measure of the change in y-coordinates over the change in x-coordinates, and a negative slope indicates a downward trend. The maximum slope, by definition, is the steepest positive slope possible.

5. How does the maximum slope relate to the steepness of a line?

The maximum slope is directly related to the steepness of a line. The steeper the line, the higher the maximum slope will be. This is because a steeper line will have a greater change in y-coordinates over a given change in x-coordinates, resulting in a larger slope.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
794
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
13
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
276
  • Calculus and Beyond Homework Help
Replies
11
Views
4K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
883
  • Calculus and Beyond Homework Help
Replies
26
Views
3K
Replies
4
Views
1K
Back
Top