A question:
There's a polynomial function of degree greater than 2, on the closed interval [a,b], such that f(a)=f(b)=1. Which of the following must be true for at least one value c in the interval [a,b]
I) f(c)=0 II) f'(c)=0 III) f''(c)=0
I am thinking that only II is true for any polynomial by Rolle's Theorem, since a polynomial satisfies all of its criteria. Are the other two options necessarily true for all polynomials stated in the problem?
There's a polynomial function of degree greater than 2, on the closed interval [a,b], such that f(a)=f(b)=1. Which of the following must be true for at least one value c in the interval [a,b]
I) f(c)=0 II) f'(c)=0 III) f''(c)=0
I am thinking that only II is true for any polynomial by Rolle's Theorem, since a polynomial satisfies all of its criteria. Are the other two options necessarily true for all polynomials stated in the problem?
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I) Nope. Proof by counter-example:
Let f(x) = -x^3 + x^2 + 1 for the interval [0, 1]
f(0) = f(1) = 1
Suppose for contradiction, that f(c) = 0 for some c in [0, 1]
=> -x^3 + x^2 + 1 = 0
=> x^3 = x^2 + 1
We know that 0 <= x^3 <= 1, and 0 <= x^2 <= 1 for x in [0, 1]
And so for any c such that f(c) = 0, c^2 must be equal to 0, and c^3 must be equal to 1.
c^3 = 1 => c = 1, but c^2 = 0 => c = 0
And so by contradiction, the first option is not necessarily true.
III) No again. Proof by counter-example:
Let f(x) = x^3 - 7x + 7 for the interval [1, 2]
f(1) = f(2) = 1
f'(x) = 3x^2 - 7
f''(x) = 3x > 0 for x in [1, 2]
You're correct about the second option. Good luck!
Let f(x) = -x^3 + x^2 + 1 for the interval [0, 1]
f(0) = f(1) = 1
Suppose for contradiction, that f(c) = 0 for some c in [0, 1]
=> -x^3 + x^2 + 1 = 0
=> x^3 = x^2 + 1
We know that 0 <= x^3 <= 1, and 0 <= x^2 <= 1 for x in [0, 1]
And so for any c such that f(c) = 0, c^2 must be equal to 0, and c^3 must be equal to 1.
c^3 = 1 => c = 1, but c^2 = 0 => c = 0
And so by contradiction, the first option is not necessarily true.
III) No again. Proof by counter-example:
Let f(x) = x^3 - 7x + 7 for the interval [1, 2]
f(1) = f(2) = 1
f'(x) = 3x^2 - 7
f''(x) = 3x > 0 for x in [1, 2]
You're correct about the second option. Good luck!
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You are correct. While all three conditions MAY hold, only the second condition MUST hold.
Take the simple example f(x) = x(x - 1)(x + 2) + 1 on the interval [0, 1]. Then f(0) = f(1) = 1, f '(c) = 0 where c ≈ 0.549 (there is exactly one critical number on this interval). f(x) > 0 for all x in [0,1] and f ''(x) > 0 for all x in [0,1].
Take the simple example f(x) = x(x - 1)(x + 2) + 1 on the interval [0, 1]. Then f(0) = f(1) = 1, f '(c) = 0 where c ≈ 0.549 (there is exactly one critical number on this interval). f(x) > 0 for all x in [0,1] and f ''(x) > 0 for all x in [0,1].
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You are absolutely correct.only second option is correct