Question

let f be a function with selected values given in the table above. which of the following statements must be true?
i. by the intermediate value theorem, there is a value c in the interval (0, 3) such that f(c) = 2.
ii. by the mean value theorem, there is a value c in the interval (0, 3) such that f(c) = 2.
iii. by the extreme value theorem, there is a value c in the interval 0, 3 such that f(c) ≤ f(x) for all x in the interval 0, 3.
the table is:

x	f(x)
1	4
2	7
3	6

Explanation:

Step1: Analyze Statement I (Intermediate Value Theorem)

The Intermediate Value Theorem (IVT) states that if a function \( f \) is continuous on \([a, b]\), and \( k \) is a number between \( f(a) \) and \( f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f(c)=k \). Here, \( f(0) = 0 \) and \( f(1)=4 \), \( f(0) = 0 \) and \( f(1) = 4 \), so the values of \( f(x) \) range from \( 0 \) (at \( x = 0 \)) to \( 7 \) (at \( x = 2 \)) and then to \( 6 \) (at \( x = 3 \)). The number \( 2 \) is between \( f(0)=0 \) and \( f(1)=4 \) (since \( 0<2<4 \)). Assuming \( f \) is continuous (a necessary condition for IVT, and since we are applying IVT, we assume continuity), by IVT, there must be a \( c \) in \((0, 3)\) (specifically in \((0, 1)\) as \( f(0)=0 \) and \( f(1)=4 \)) such that \( f(c) = 2 \). So Statement I is a candidate.

Step2: Analyze Statement II (Mean Value Theorem)

The Mean Value Theorem (MVT) states that if a function \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists a \( c \) in \((a, b)\) such that \( f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\). Here, \( a = 0 \), \( b = 3 \), \( f(0)=0 \), \( f(3)=6 \). So \(\frac{f(3)-f(0)}{3 - 0}=\frac{6 - 0}{3}=2\). But MVT requires the function to be differentiable on \((0, 3)\) and continuous on \([0, 3]\). However, we don't know if \( f \) is differentiable. The problem only gives a table of values, not information about differentiability. So we cannot be sure that MVT applies, so Statement II may not be true.

Step3: Analyze Statement III (Extreme Value Theorem)

The Extreme Value Theorem (EVT) states that if a function \( f \) is continuous on a closed interval \([a, b]\), then \( f \) attains a maximum and a minimum on \([a, b]\). But Statement III says "there is a value \( c \) in \([0, 3]\) such that \( f(c)\leq f(x) \) for all \( x \) in \([0, 3]\)", which is saying \( f(c) \) is a minimum. However, \( f(0)=0 \), and we need to check if \( 0 \) is the minimum. But \( f(0)=0 \), and we don't know if there are any values of \( f(x) \) less than \( 0 \) in \([0, 3]\). But from the table, the smallest value we see is \( 0 \) at \( x = 0 \). However, the Extreme Value Theorem guarantees the existence of a minimum (and maximum) on a closed interval for continuous functions. But the statement says "for all \( x \) in \([0, 3]\)", but we only have discrete points. Also, the way it's phrased: " \( f(c)\leq f(x) \) for all \( x \)" – the minimum value of \( f(x) \) in the table is \( 0 \) (at \( x = 0 \)), but does that mean \( f(0)\leq f(x) \) for all \( x \)? \( f(1)=4\geq0 \), \( f(2)=7\geq0 \), \( f(3)=6\geq0 \), so \( f(0) = 0 \) is less than or equal to all other \( f(x) \) values in the table. But the Extreme Value Theorem is about the existence of a minimum (and maximum) on a closed interval for continuous functions. However, the statement is not exactly the EVT's conclusion (EVT says there exists a minimum and a maximum, not that a particular point is the minimum for all \( x \) in the way stated, and also we don't know if the function is continuous outside the table points). Also, the key is that Statement I is about IVT which, given the values (assuming continuity), must hold, while Statement II requires differentiability (not given) and Statement III is misapplying EVT (EVT says there is a minimum, but the way it's stated is a bit off, but more importantly, Statement I is more straightforward with the given table).

Answer:

I. By the Intermediate Value Theorem, there is a value \( c \) in the interval \((0, 3)\) such that \( f(c) = 2 \)