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CALCULUS HELP (1 pt) Consider the function f(x) = 2 x^3 - 5 x on the interval [ -2 , 2 ]. Find the average or mean slope of the function on this interval. By the Mean Value Theorem, we know there exists at least one c in the open interval ( -2 , 2 ) such that f'( c) is equal to this mean slope. For this problem, there are two values of c that work. The smaller one is= and the larger one is= i found the mean slope=3 but cannot get the values where f'(c) is equal to 3.... tried setting the derivative equal to three but that doesnt yield the right answer, or im not working it out right....please

Mathematics
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u mena find the equation of tangent line?
|dw:1353882845431:dw|
i dont think so... i need to find where f' is = to the mean slope(3) on the interval (-2,2)..tried setting f' =3 but with no luck

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Other answers:

ok it means the same ok so use the formula above
find derivative first
what did u get
6x^2-5
now calculate |dw:1353883173646:dw|
its 3
set the derivative=3 solve for x
awesome, found my mistake...thank you man
what did u get?
-1.1545, 1.1545
yup u got it yup
anymore question>
yes .. (1 pt) Consider the function f(x) = 1/x on the interval [ 1 , 10 ]. (A) Find the average or mean slope of the function on this interval. - i found it to be -.1 (B) By the Mean Value Theorem, we know there exists a c in the open interval ( 1 , 10 ) such that f'( c) is equal to this mean slope. Find all values of c that work and list them (separated by commas) in the box below. - i set -1/x^2 = -.1 and got -3.1622 and 3.1622 but it doesnt accept those as answers
yup
correct
ok i had to use only the positive one since it was in the interval given........thank you for your help..... have an awesome day !

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