Question

How do you verify the intermediate value theorem over the interval [0,3], and find the c that is guaranteed by the theorem such that f(c)=0 where `f(x)=x^2-6x+8`?

Answer 1

Intermediate Value theorem is verified so, there exist `c in [0,3]`
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such that `f (c)=0`
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`c=2`

Explanation:

`f` is continuous over `[0,3]` and
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`f (0)=0^2-6 (0)+8=8`
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`f (3)=3^2-6 (3)+8 = 9-18+8=-1`
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Intermediate Value theorem is defined over `[0,3]` for `f (x)`
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Because
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`f` is continuous over `[0,3]` and `f (3)< f (c) < f (0)`
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then there exist `0 < c < 3` such that `f (c)=0" "`
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`c = ??`
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`f (c)=0`
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`rArrc^2 -6c +8 = 0`
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`rArr c^2 -6c +8 + 9 -9 =0`
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`rArr c^2 - 6c +9 -9 +8 = 0`
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`rArr (c^2 -2 (3)c +3^2)-9+8=0`
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Factorization in the previous step is determined by applying :
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`(a-b)^2=a^2-2ab+b^2`
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`rArr (c-3)^2-1=0`
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`rArr (c-3)^2-1^2=0`
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Factorization here is determined by applying the followingbidentiyy:
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`a^2-b^2=(a-b)(a+b)`
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`rArr ((c-3)-1)((c-3)+1)=0`
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`rArr (c-4)(c-2)=0`
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`rArrc-4=0rArrc=4" "`Rejected since`" "4!in [0,3]`
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Or
`rArrc-2=0rArrc=2" "`Accepted since`" "2in [0,3]`