Instantaneous and Average Rate of Change of f(x) | Interval (0,3)

[pyrosilver]

Homework Statement

Find any and all points inside the interval (0,3) where the instantaneous rate of change of f equals the average rate of change of f over the interval [0,3], for the equation

f(x) = 4x^2 - x^3

Homework Equations

The Attempt at a Solution

Not really sure how to do this. How do I find the average rate of change over that interval? I was thinking take the derivative:

f'(x) = 8x - 3x^2

plugging in 0 and 3 to get f'(0) = 0, and f'(3) = -3. So the average rate of change would be
-1.5?

And then saying

-1.5 = = 8x - 3x^2? I don't know, gah. How do I solve this problem?

 

[theJorge551]

Based on how I see things, you're quite nearly there! You've just ended up with a quadratic equation, now what do you do?

 

[JonF]

Derivatives are the instantaneous rate of change. Your book should have a formula (a very familiar looking formula) for average rates of change.

 

[VietDao29]

Homework Statement

Find any and all points inside the interval (0,3) where the instantaneous rate of change of f equals the average rate of change of f over the interval [0,3], for the equation

f(x) = 4x^2 - x^3

Homework Equations

The Attempt at a Solution

Not really sure how to do this. How do I find the average rate of change over that interval? I was thinking take the derivative:

f'(x) = 8x - 3x^2

plugging in 0 and 3 to get f'(0) = 0, and f'(3) = -3. So the average rate of change would be
-1.5?

And then saying

-1.5 = = 8x - 3x^2? I don't know, gah. How do I solve this problem?

No, what you've written is not the "average rate of change". The average rate of change of a function f over the interval [a; b] should be calculated as follow:

\frac{f(b) - f(a)}{b - a}

Hopefully, you can go from there, right? :)