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How would I solve that? thanks, I will choose best answer.
How would I solve that? thanks, I will choose best answer.
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The question is kind of silly. If f(x) is everywhere increasing, then f(x2) > f(x1) for any x2 > x1 in the domain. So the MVT is completely unnecessary. Its use also requires us to make some assumptions about f.
Suppose that we take for granted that f is continuous on [-1,0] and differentiable on (-1, 0). As f is increasing, we must have f '(x) > 0 for all x in (-1,0).
But, the MVT guarantees that for some c in (-1, 0)
f '(c) = (f(-1) - f(0))/(-1 - 0) = (1/2)(-1) = -1/2.
This is a contradiction. So it must be that no such function can exist.
The second question is just a repeat of the first. Suppose f '(-1) = 1/2 and f '(0) = 0 and that f ''(x) > 0 on (-1,0). We get the same issue arising.
The MVT tells us that there exists b in (-1, 0) such that
f ''(b) = ((f '(-1) - f '(0))/(-1 - 0) = -1/2
in direct contradiction to f '' (x) > 0.
I guess the second question is just to remind you that the relationship between f ' and f is the same as that between f '' and f '.
Suppose that we take for granted that f is continuous on [-1,0] and differentiable on (-1, 0). As f is increasing, we must have f '(x) > 0 for all x in (-1,0).
But, the MVT guarantees that for some c in (-1, 0)
f '(c) = (f(-1) - f(0))/(-1 - 0) = (1/2)(-1) = -1/2.
This is a contradiction. So it must be that no such function can exist.
The second question is just a repeat of the first. Suppose f '(-1) = 1/2 and f '(0) = 0 and that f ''(x) > 0 on (-1,0). We get the same issue arising.
The MVT tells us that there exists b in (-1, 0) such that
f ''(b) = ((f '(-1) - f '(0))/(-1 - 0) = -1/2
in direct contradiction to f '' (x) > 0.
I guess the second question is just to remind you that the relationship between f ' and f is the same as that between f '' and f '.