Question

let /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 $f(c)=\frac{4}{9}$ 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 \(f(b)\) (\(f(a)
eq f(b)\)), 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 \(f(6)=3\). The value \(k = 1\) is between \(-1\) and \(3\) (\(-1<1<3\)).

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 basis to say \(f(0)=0\).
• Option B: \(f^{\prime}(c)=\frac{4}{9}\) is related to the Mean Value Theorem (\(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}=\frac{3-(-1)}{6-(-3)}=\frac{4}{9}\) when \(f\) is differentiable on \((-3,6)\) and continuous on \([-3,6]\)), not the Intermediate Value Theorem.
• Option C: The Extreme Value Theorem (if \(f\) is continuous on \([a,b]\), then \(f\) has a maximum \(M\) and a minimum \(m\) on \([a,b]\) such that \(m\leq f(x)\leq M\) for all \(x\in[a,b]\)) is needed to guarantee bounds for all \(x\) in the interval, not the Intermediate Value Theorem.
• Option D: Since \(f\) is continuous on \([-3,6]\), \(f(-3)=-1\), \(f(6)=3\), and \(1\) is between \(-1\) and \(3\), 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 where the function is given to be continuous (\([-3,6]\) in the problem statement).

Answer:

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