Suppose that y=f(x) is a continous function on x∈[-1,3], and f(-1)=4, f(3)=7. We can conclude that
a) there exists c∈(-1,3) such that f(c)=0
b) there exists c∈(-1,3) such that f(c)=5
c) there exists c∈(4,7) such that f(c)=0
d) there exists c∈(4,7) such that f(c)=5
e) there exists c∈(-1,1) such that f(c)=3
By the intermediate value theorem, we have that "\\exist c \\in (-1,3) \\ni f(c) = 5" . Option (b) is therefore the answer
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