How do you verify the intermediate value theorem over the interval [0,5], and find the c that is guaranteed by the theorem such that f(c)=11 where #f(x)=x^2+x-1#?

1 Answer
Narad T.
Sep 4, 2017

The value of #c=3#

Explanation:

The Intermediate Value Theorem states that if #f(x)# is a continuous function on the interval #[a,b]# and #N in (f(a), f(b)) #, then there exists #c in [a,b]# such that #f(c)=N#

Here,

#f(x)=x^2+x-1# is continuous on #RR# as it is a polynomial function.

The interval is #I= [0,5]#

#f(0)=-1#

#f(5)=25+5-1=29#

#f(x) in [ -1, 29]#

Then,

#f(c)=11# , #=># #11 in [-1,29]#

#EE c in [0,5]# such that #f(c)=11#

Therefore,

#c^2+c-1=11#, #=>#, #c^2+c-12=0#

#(c+4)(c-3)=0#

#c=-4# and #c=3#

#c=3# and #c in [0,5]#

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