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#?

1 Answer
Nov 1, 2017

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]#