The table has
x values 2,3,4,5 and
f(x) as 1, 14,20, 31
The statements A is true Intermediate value theorem, B is false mean value theorem, C is true extreme value theorem and D is true.
Given that,
The table has
x values 2,3,4,5 and
f(x) as 1, 14,20, 31
The function f is continuous.
A is true, From the figure.
Intermediate value theorem is let [a,b]be a closed and bounded intervals and a function f:[a,b]→R be continuous on [a,b]. If f(a)≠f(b) then f attains every value between f(a) and f(b) at least once in the open interval (a,b).
B is false because, mean value theorem, Let a function f:[a,b]→R be such that,
1. f is continuous on[a,b] and
2. f is differentiable at every point on (a,b).
Then there exist at least a point c in (a,b) such that f'(c)=(f(b)-f(a))/b-a
In the B part, the differentiability is not given do mean value theorem can be applied.
C is true because the extreme value theorem, if a real-valued function f is continuous on the closed interval [a,b] then f attains a maximum and a minimum each at least once such that ∈ number c and d in[a,b] such that f(d)≤f(x)≤f(c)∀ a∈[a,b].
D is true.
Therefore, The statements A is true, B is false, C is true and D is true.
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