I will reword your "proofs" so that they will be correct. However, I must use my version of the
Intermediate Value Theorem because your still makes no sense to me (for instance, 2. appears to be the conclusion of the
Extreme Value Theorem...).
Quote:
Originally posted by frozenstellar
i've attempted two questions i had, and if someone could post any feedback, it'd be great.
1. If f(x) = x^3 - 5x^2 + 7x - 9, prove there is a real number c such that f(c) = 100.
let g(x) = x^3 - 5x^2 + 7x - 109 where g(c) = 0.
when x = 0, g(0) = -109
when x = 10, g(10) = 461
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This is the part you got right. You picked a good interval to work with and this was something you needed to do if you wanted to use the
Intermediate Value Theorem. However, as
Yakk pointed out, you made a false conclusion and then tried to build off of it. The rest of your proof should have looked something like this:
Now, because g(0) < 0, g(10) > 0, and g(x) is continuous, we know, from the
Intermediate Value Theroem, that there exists a c in [0,10] such that g(c) = 0.
But g(c) = c^3 - 5c^2 + 7c - 109.
So, c^3 - 5c^2 + 7c - 109 = 0,
or c^3 - 5c^2 + 7c - 9 = 100.
QED.
The key here was to
use the
Intermediate Value Theorem to prove your conclusion.
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Your second part was similarly flawed but I'll leave it to you to correct it. I'm sure you know how!