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ƒ(x) ≤ g(x) ≤ h(x) for all x and if the limit of ƒ(x) as x→c = L and the limit of h(x) as x→c = L, then the
limit of g(x) as x→c = L ✔Correct Answer--Squeeze Theorem
If ƒ is continuous on [a,b] and k is a value between ƒ(a) and ƒ(b), then there is at least one c in [a,b]
such that ƒ(c) = k ✔Correct Answer--Intermediate Value Theorem
If ƒ(x) is continuous on [a,b], then ƒ(x) has a max and a min ✔Correct Answer--Extreme Value
Theorem
The absolue extrema of a function continuous on [a,b] will occur at either the critical numbers or at
the endpoints ✔Correct Answer--Absolute Extrema Theorem
If ƒ(x) is continuous on [a,b] and differentiable on (a,b) and ƒ(a) = ƒ(b), then there is c ∈ (a,b) such
that ƒ'(c) = 0 ✔Correct Answer--Rolle's Theorem
If ƒ(x) is continuous on [a,b] and differentiable on (a,b), then there is a c ∈ (a,b) such that ƒ'(c) = [ƒ(b)
- ƒ(a)] ÷ (b - a) ✔Correct Answer--Mean Value Theorem
If the limit of ƒ(x)/g(x) as x→a is indeterminent, then you can use the limit of ƒ'(x)/g'(x) as x→a to
find the limit ✔Correct Answer--L' Hôpital's Rule
If ƒ(x) is continuous on [a,b], then ∫(from a to b) ƒ(x) dx = Ƒ(b) - Ƒ(a) ✔Correct Answer--First
Fundamental Theorem of Calculus
If ƒ(x) is continuous on [a,b], then there is a c such that ∫(from a to b) ƒ(x) dx = ƒ(c)(b - a) ✔Correct
Answer--Mean Value Theorem for Integrals
d/dx ∫(from x to a) ƒ(t) dt = d/dx [ ƒ(x) - ƒ(a) ] = ƒ(x) ✔Correct Answer--Second Fundamental
Theorem of Calculus