What is the maximum slope for this line?

[zJakeAdam]

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?

 

[Undoubtedly0]

Greetings and welcome!

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:

$$slope = \frac{y_2 - y_1}{x_2 - x_1}$$.

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.

 

[zJakeAdam]

Greetings and welcome!



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

$$slope = \frac{y_2 - y_1}{x_2 - x_1}$$.

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.

 

[Undoubtedly0]

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?