Question

let h be a continuous function on the closed interval 1,6. a few values of h are given in this table:

x	1	3	4	6

which intervals must contain a solution to h(x) = 0?

choose all answers that apply:

a 1,3

b 3,4

c 4,6

d none of the above

Explanation:

To determine which intervals must contain a solution to \( h(x) = 0 \), we use the Intermediate Value Theorem (IVT). The IVT states that if a function \( h \) is continuous on a closed interval \([a, b]\), and \( k \) is a value between \( h(a) \) and \( h(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( h(c) = k \). In this case, we want to find intervals where \( 0 \) is between \( h(a) \) and \( h(b) \).

Step 1: Analyze Interval \([1, 3]\)

For the interval \([1, 3]\), we have \( h(1) = -5 \) and \( h(3) = -2 \). Both values are negative, so \( 0 \) is not between \( -5 \) and \( -2 \). Thus, by IVT, there is no guarantee of a solution in \([1, 3]\).

Step 2: Analyze Interval \([3, 4]\)

For the interval \([3, 4]\), we have \( h(3) = -2 \) and \( h(4) = 2 \). Here, \( -2 < 0 < 2 \), so \( 0 \) is between \( h(3) \) and \( h(4) \). Since \( h \) is continuous on \([3, 4]\), by IVT, there must be some \( c \) in \((3, 4)\) such that \( h(c) = 0 \). So, \([3, 4]\) must contain a solution.

Step 3: Analyze Interval \([4, 6]\)

For the interval \([4, 6]\), we have \( h(4) = 2 \) and \( h(6) = -5 \). Here, \( -5 < 0 < 2 \), so \( 0 \) is between \( h(4) \) and \( h(6) \). Since \( h \) is continuous on \([4, 6]\), by IVT, there must be some \( c \) in \((4, 6)\) such that \( h(c) = 0 \). So, \([4, 6]\) must contain a solution.

Answer:

B. \([3, 4]\)
C. \([4, 6]\)