If f is continuous on [-4,4] such that f(-4)=11 and f(4)=-11 then
b) lim as x approaches 2 f(x)=8
c) There is at least one c [-4,4] such that f(c)=8
d) lim as x approaches 3 f(x) = lim as x approaches -3
e) It is possible that f is not defined at x=0
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Let's break down the question:
We know that f(x) is continuous between x = -4 and x = 4. That means there are no breaks in the line if we put it on a graph.
If there are no breaks in the line, and we know that the value of the function goes from f(x) = 11 to f(x) = -11, then we know that every value betwen 11 and -11 has to be hit by f(x) between x = -4 and x = 4. Otherwise, you'd have to jump over certain numbers, and there would be a discontinuity! This is called the "Intermediate Value Theorem."
Therefore, we know that there must exist an x between -4 and 4 that will map to any value between -11 and 11 for f(x).
More precisely, for any value z such that -11 ≤ z ≤ 11, there must exist a number c where -4 ≤ c ≤ 4 and f(c) = z.
(C) is the option that says this with the specific example that z = 8. Because we do not know the function, we cannot figure out what "c" is or even if there are more than just one value possible for "c." We do know, though, that there is at least one value for which f(c) = 8.
Intermediate value problem! Keep your eye out for these.We don't know any specifics about what the function does between -4 and 4, except that it somehow goes from 11 down to -11.
So f(x) has to take on all values between 11 and -11 somewhere on the interval from -4 to 4. In this case, choice C asks if f will ever be 8. We know it must be. Go through the other answers to make sure you know why they're not correct.
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