Question

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1. let ( f ) be a continuous function on the closed interval ( -3,6 ). if ( f(-3)=-1 ) and ( f(6)=3 ), then the intermediate value theorem guarantees that

(a) ( f(0)=0 )
(b) ( f^{prime}(c)=\frac{4}{9} ) for at least one ( c ) between ( -3 ) and 6
(c) ( -1 leq f(x) leq 3 ) for all ( x ) between ( -3 ) and 6
(d) ( f(c)=1 ) for at least one ( c ) between ( -3 ) and 6
(e) ( f(c)=0 ) for at least one ( c ) between ( -1 ) and 3

Explanation:

Step1: Recall the Intermediate Value Theorem

The Intermediate Value Theorem states that if \(y = f(x)\) is continuous on a closed interval \([a,b]\), and \(k\) is a number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\in(a,b)\) such that \(f(c)=k\). Here, \(a=-3\), \(b = 6\), \(f(-3)=-1\), and \(f(6)=3\).

Step2: Analyze each option

• Option A:

The Intermediate Value Theorem does not guarantee that \(f(0)=0\). Just because the function is continuous on \([-3,6]\) and \(f(-3)=-1\), \(f(6)=3\), there is no information about the value of the function at \(x = 0\) from the given data for the Intermediate Value Theorem.

• Option B:

\(f^{\prime}(c)=\frac{4}{9}\) is related to the Mean Value Theorem (\(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}=\frac{3-(-1)}{6-(-3)}=\frac{4}{9}\) if \(f\) is differentiable on \((-3,6)\)). But the problem only states that \(f\) is continuous, not differentiable. So, the Intermediate Value Theorem does not guarantee this.

• Option C:

The Intermediate Value Theorem does not give information about the bounds of the function for all \(x\) in the interval. It only gives information about values that the function must take, not about the maximum and minimum values of the function on the interval.

• Option D:

Since \(f\) is continuous on \([-3,6]\), and \(k = 1\) is between \(f(-3)=-1\) and \(f(6)=3\) (\(-1<1<3\)). By the Intermediate Value Theorem, there exists at least one \(c\in(-3,6)\) such that \(f(c)=1\).

• Option E:

The interval \([-1,3]\) is not the interval \([-3,6]\) on which we know the function is continuous. The Intermediate Value Theorem is applied on the interval \([a,b]=[-3,6]\) for values between \(f(-3)\) and \(f(6)\).

Answer:

D. \(f(c) = 1\) for at least one \(c\) between \(-3\) and \(6\)