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Mathematics 21 Online
OpenStudy (anonymous):

How do you know what is the rate of change of the steepest section of an exponential graph?

OpenStudy (anonymous):

Differentiate the exponential graphs equation- at the x point substitute the last value of the x that counts in the domain of the graph (that looks the steepest).

OpenStudy (anonymous):

I'm not supposed to use differentiation - is there another way though? The picture I've attached is my exponential function that I'm talking about, with those domain restrictions.

OpenStudy (anonymous):

Well rate of change refers to \(\dfrac{dy}{dx}\)...

OpenStudy (anonymous):

Anyway can you draw the graph?

OpenStudy (anonymous):

I know lol - but I'm taking advanced functions right now and we're not supposed to use that yet. We can use the different quotient and other things but not that

OpenStudy (anonymous):

Alright.

OpenStudy (anonymous):

It's the graph in black...

OpenStudy (anonymous):

So there are no points to it?

OpenStudy (anonymous):

Like on the x and y axis- no numbers?

OpenStudy (anonymous):

No, we just use that equation, and from the equation and the graph, determine where the steepest point is

OpenStudy (anonymous):

So you need the coordinates of that point?

OpenStudy (anonymous):

Yeah, but my question is how do I get those coordinates/point?

OpenStudy (anonymous):

Differentiate the exponential graphs equation- at the x point substitute the last value of the x that counts in the domain of the graph (that looks the steepest). ^You say this, so my last value of x is 28.3 (as mentioned in the restriction of the domain), so do I use this?

OpenStudy (anonymous):

And after differentiating the exponential equation, and substitute 28.3, I find a value, but what is that value used for?

OpenStudy (anonymous):

That's the gradient at that point- rate of change.

OpenStudy (anonymous):

Ohh, so that's the value that I'm supposed to find then! Got it now, thank you!

OpenStudy (anonymous):

No worries :)

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