I'll post what I have already solved, but I'm not sure if there's more to the question.

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I'll post what I have already solved, but I'm not sure if there's more to the question.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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f(0) = (0)^3 = 0 f(1) = (1)^3 = 1 \[\frac{ 1-0 }{ 1-0 }=\frac{1}{1}=1\] The derivative of the function is 3x^2 3x^2 = 1 x^2= 1/3 x = sqrt(1/3) or -sqrt(1/3) Since only sqrt(1/3) is found in [0,1], then it is the only answer.
Is there anything else this question is asking me to do? (It's just worded confusingly to me)

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do you have to define how the mean value thrm applies to it? i think it has to do with being continuous along the interval
other than that, you did well at finding x=c that has a slope equal to the slope between x=a and x=b
What do mean by define how it applies?
what are the required conditions for the MVT to be applicable?
The function must be continuous and differentiable
..." if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b)" ... http://www.sosmath.com/calculus/diff/der11/der11.html correct, and since this function is defined, continuous, and diffy-able then yada yada
one of the common examples of a fail for this is f(x) = 1/x, on the interval (-1,1)
a slope can be defined across the interval, but the function has no value in the interval that matches that slope ... it is not continuous or diffy-able at x=0 within the interval
Ah okay. Could you interpret this by chance? ... "Find any tangent lines to the graph of f that are parallel to the secant line."
I could post the problem if it would help
more info usually helps, but the idea is to find the slope ofthe secant line in order to evaluate the derivative tha tmatches it
Is there a formula to finding the slope of the secant line?
Oh nevermind! It's 2/3. I've already solved for it.
:) f(b)-f(a) ------- b-a
So how would I find the parallel lines?
parallel lines have the same slope ....
So it would be any function with the slope 2/3?
the derivative of the function (if one is given) define the slope at any single point
any value x=c of which f'(c) = 2/3
this is just another wording of the MVT
Okay thank you!
youre welcome was stuck in OS limbo for a bit :/

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