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Is There A Number That Is Exactly 1 More Than Its Cube, Proof Verification

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So, Im studying limits, and got to the incredible cool intermediate value theorem now, and found this question in my book. Even though he doesnt ask for THE number, he wants me to show that it exists, so i tried:(displaystyle x = x³ + 1)now, here is where the problem stats, i can form two different functions, (displaystyle f(x) = x³ + 1 – x)(displaystyle g(x) = -x³ – 1 + x)By testing, however, the answer is the same, x is in (displaystyle (-1,4 , -1,3)). But when I actually look at the results, I notice that:(displaystyle f(-1,3) = 0,103 )(displaystyle g(-1,3) = -0,103)(displaystyle f(-1,4) = -0,344)(displaystyle g(-1,4) = 0,344)Even though in this case the function I choose wont matter, since I will get to the answer, Should I have picked one over another?
So, Im studying limits, and got to the incredible cool intermediate value theorem now, and found this question in my book. Even though he doesnt ask for THE number, he wants me to show that it exists, so i tried:(displaystyle x = x³ + 1)(displaystyle f(x) = x³ + 1 – x)Even though in this case the function I choose wont matter, since I will get to the answer, Should I have picked one over another?
No, just note that (displaystyle f(-2)That is sufficient to show that (displaystyle f(x)=0) for some (displaystyle x.)

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For your function (displaystyle displaystyle y = x^3 – x + 1), note that when (displaystyle displaystyle x = -2, y = -5), and when (displaystyle displaystyle x = 1, y = 1).Since the function”s value has gone from something negative to something positive, it must have crossed the (displaystyle displaystyle x) axis somewhere at least once.So a solution exists.Edit: Crap, just realised that Plato posted the exact same thing – serves me right for not reading the responses more carefully haha.

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