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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0f(0) = (0)^3 = 0 f(1) = (1)^3 = 1 \[\frac{ 10 }{ 10 }=\frac{1}{1}=1\] 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is there anything else this question is asking me to do? (It's just worded confusingly to me)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3do you have to define how the mean value thrm applies to it? i think it has to do with being continuous along the interval

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3other than that, you did well at finding x=c that has a slope equal to the slope between x=a and x=b

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What do mean by define how it applies?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3what are the required conditions for the MVT to be applicable?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The function must be continuous and differentiable

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3..." 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 diffyable then yada yada

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3one of the common examples of a fail for this is f(x) = 1/x, on the interval (1,1)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3a 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 diffyable at x=0 within the interval

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah okay. Could you interpret this by chance? ... "Find any tangent lines to the graph of f that are parallel to the secant line."

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I could post the problem if it would help

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3more info usually helps, but the idea is to find the slope ofthe secant line in order to evaluate the derivative tha tmatches it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is there a formula to finding the slope of the secant line?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh nevermind! It's 2/3. I've already solved for it.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3:) f(b)f(a)  ba

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So how would I find the parallel lines?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3parallel lines have the same slope ....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So it would be any function with the slope 2/3?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3the derivative of the function (if one is given) define the slope at any single point

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3any value x=c of which f'(c) = 2/3

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3this is just another wording of the MVT

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3youre welcome was stuck in OS limbo for a bit :/
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