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Also, at what point does the curve have said slope?
Define "smallest". Is 0 SMALLER than -3, for example?
I think it means "smallest" as lowest in its context.
That doesn't mean anything. The choices are: Closest to zero. Greatest possible negative.
Greatest possible negative.
Okay, how shall we proceed? Do we get to use the calculus?
Yep, I'm in Calc 1, so that's what I'd prefer.
Fine. This is an interesting question because we don't much care about the original function. We are much more interested in the derivative. Please find the 1st Derivative and see if that helps answer the question.
I got Y prime=3x^2-4 for the derivative.
That's correct. No what? Reading the problem statement, we want to minimize THIS function. We want to know about the SLOPE of the derivative - not the value of the derivative. We're going to need the 2nd derivative. Pleese provide.
F double prime= 6x. But i'm guessing that won't do me any good either, so the third derivative is 6. Is my math right?
Why won't that do you any good? You want to know where a function has it's minimum. Isn't that the job of a derivative? If I just handed you this problem, could you solve it? Find the minimum value of f(x) = 3x^2 - 4 What would you do?
Plug in 0 for x?
Or find the derivative?
Very good. The ONLY point of confusion on this problem, and hardly anyone gets it on the first try, is that we are trying to minimize a function that we already know in our heads is a 1st derivative. If we can get over that, we'll be fine. Original Problem Statement: What is the smallest slope on the curve y=x^3-4x+1? Rephrase (specific): What is the minimum value of y'? Rephrase (Generic): What is the minimum value of f(x) = 3x^2 - 4?
So the minimum value of y' is 6x?
6x is not a value. I'm trying to get you to forget that it is y'. The minimum value of 3x^2 - 4 is -4. If you would like to find the derivative of f(x) = 3x^2 - 4 to determine that x = 0 is where the minimum value exists, then feel free to do that.
So to find the minimum value of f(x)= 3x^2-4, all I have to do is plug in a zero into that equation?
If you happen to know that x = 0 is the right place, yes. If you took enough analytic geometry before calculus, as it se3ems you have, this should be relatively obvious.
I'm sorry, I'm just not understanding why zero is the right place?
So, maybe you haven't had enough analytic geometry? Faced with the parabola, y = 3x^2 - 4, where is the vertex? Or, like you did up above, find where the 1st derivative is zero. Either way, you get x = 0 as the right spot.
Oh okay I understand now.