# 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?

Nov 1, 2017

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

#### Explanation:

$f$ is continuous over $\left[0 , 3\right]$ and
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$f \left(0\right) = {0}^{2} - 6 \left(0\right) + 8 = 8$
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$f \left(3\right) = {3}^{2} - 6 \left(3\right) + 8 = 9 - 18 + 8 = - 1$
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Intermediate Value theorem is defined over $\left[0 , 3\right]$ for $f \left(x\right)$
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Because
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$f$ is continuous over $\left[0 , 3\right]$ and $f \left(3\right) < f \left(c\right) < f \left(0\right)$
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then there exist $0 < c < 3$ such that $f \left(c\right) = 0 \text{ }$
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c = ??
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$f \left(c\right) = 0$
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$\Rightarrow {c}^{2} - 6 c + 8 = 0$
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$\Rightarrow {c}^{2} - 6 c + 8 + 9 - 9 = 0$
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$\Rightarrow {c}^{2} - 6 c + 9 - 9 + 8 = 0$
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$\Rightarrow \left({c}^{2} - 2 \left(3\right) c + {3}^{2}\right) - 9 + 8 = 0$
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Factorization in the previous step is determined by applying :
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${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$
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$\Rightarrow {\left(c - 3\right)}^{2} - 1 = 0$
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$\Rightarrow {\left(c - 3\right)}^{2} - {1}^{2} = 0$
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Factorization here is determined by applying the followingbidentiyy:
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${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
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$\Rightarrow \left(\left(c - 3\right) - 1\right) \left(\left(c - 3\right) + 1\right) = 0$
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$\Rightarrow \left(c - 4\right) \left(c - 2\right) = 0$
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$\Rightarrow c - 4 = 0 \Rightarrow c = 4 \text{ }$Rejected since$\text{ } 4 \notin \left[0 , 3\right]$
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Or
$\Rightarrow c - 2 = 0 \Rightarrow c = 2 \text{ }$Accepted since$\text{ } 2 \in \left[0 , 3\right]$