What is the maximum value of the first derivative of f(x)=3x^2-x^3?
1) find the first derivative of f(x). 2) Treat f '(x) as a new function. Go through the necessary steps to identify the max value of f '(x), again applying the derivative. Please show y our work.
The first derivative I got was 6x-3x^2 The second derivative I got was 6-6x
Let's review: Suppose I give you function g(x) and ask you to find the max of this function. What would you do to g(x)? List the steps.
Find the first derivative, then the second and set it equal to 0 and solve for x.
I need help, I thought I knew the right way to do it but I was clearly wrong
Yes, IF you meant "set the first derivative = to 0." You have found the derivative of the given function, obtaining 6x-3x^2. Now pretend that this is g(x): g(x)=6x-3x^2. Again, I ask you how to determine the max value of g(x). Please go thru the actual steps, sharing your work.
I thought you set the second derivative to 0
g(x) = 6x-3x^2 g '(x) = ? Set g '(x) = 0 and solve for x. Your result?
Actually, that's exactly what I'm asking you to do (immediately above).
Well I did that and got x=1 but the correct answer is x=3
Give me a second, please; I want to try this on paper.
You say you've found x=1? Are you 100% sure that "the correct answer is x=3?"
Try taking the derivative of 6-6x. Your result?
Yes. I got the answer from the answer key to this book.
Are you sure that the label for the answer is "x="?
The answer says D. The choices where a) 0 b) 1 c) 2 d) 3 e) 4
I note that none of these have labels. If you say "the answer is 3," you are speaking about the value of the first derivative of the given function. In other words, you are to find the value of f '(x) = 6x -3x^2 at x=1. Try it, please.
Okay I got the right answer! Thanks!
so, you see, it was not correct to label the max value of the deriv. of f(x) "x=3". Rather, the correct answer would be f '(1)=3.
Yeah. I got that already.
Glad you're making progress.
\[A solution and an associated plot is attached.\]
Join our real-time social learning platform and learn together with your friends!