Hi I need some help with steps on this problem, I know a slope has to be found from 2 points but after that I don't know what, here is the link , its problem 16
http://www.baruch.cuny.edu/sacc/documents/2205AdditionalProblemsfortheFEM.pdf
http://www.baruch.cuny.edu/sacc/documents/2205AdditionalProblemsfortheFEM.pdf
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x ....... f(x)
2.5 ... 31.25
2.8 ... 39.2
3.0 ... 45
3.1 ... 48.05
Function f is differentiable, and both f and f' are strictly increasing on the interval
0 ≤ x ≤ 5
QUESTION: Which of the following could be the value of f'(3) ?
A) 20
B) 27.5
C) 29
D) 30
E) 48.05
By Mean Value Theorem, there is some value c where 2.8 < c < 3.0 such that
f'(c) = (f(3.0) − f(2.8)) / (3.0 − 2.8) = (45 − 39.2) / 0.2 = 29
By Mean Value Theorem, there is some value d where 3.0 < d < 3.1 such that
f'(d) = (f(3.1) − f(3.0)) / (3.1 − 3.0) = (48.05 − 45) / 0.1 = 30.5
Since f'(x) is strictly increasing on interval 0 ≤ x ≤ 5, and 0 < c < 3.0 < d < 5 then
f'(c) < f'(3) < f'(d)
29 < f'(3) < 30.5
The only solution that is within the interval is: D) 30
2.5 ... 31.25
2.8 ... 39.2
3.0 ... 45
3.1 ... 48.05
Function f is differentiable, and both f and f' are strictly increasing on the interval
0 ≤ x ≤ 5
QUESTION: Which of the following could be the value of f'(3) ?
A) 20
B) 27.5
C) 29
D) 30
E) 48.05
By Mean Value Theorem, there is some value c where 2.8 < c < 3.0 such that
f'(c) = (f(3.0) − f(2.8)) / (3.0 − 2.8) = (45 − 39.2) / 0.2 = 29
By Mean Value Theorem, there is some value d where 3.0 < d < 3.1 such that
f'(d) = (f(3.1) − f(3.0)) / (3.1 − 3.0) = (48.05 − 45) / 0.1 = 30.5
Since f'(x) is strictly increasing on interval 0 ≤ x ≤ 5, and 0 < c < 3.0 < d < 5 then
f'(c) < f'(3) < f'(d)
29 < f'(3) < 30.5
The only solution that is within the interval is: D) 30
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Find slopes of secant lines. x = 2.8 to 3.0 and 3.0 to 3.1. Recall the initial statement: f and f ' are strictly increasing.