Question

let ( h ) be a continuous function on the closed interval ( -3, 4 ), where ( h(-3) = -1 ) and ( h(4) = 2 ).
which of the following is guaranteed by the intermediate value theorem?
choose 1 answer:
( \boldsymbol{\text{a}} ) ( h(c) = 1 ) for at least one ( c ) between ( -1 ) and ( 2 )
( \boldsymbol{\text{b}} ) ( h(c) = -2 ) for at least one ( c ) between ( -3 ) and ( 4 )
( \boldsymbol{\text{c}} ) ( h(c) = -2 ) for at least one ( c ) between ( -1 ) and ( 2 )
( \boldsymbol{\text{d}} ) ( h(c) = 1 ) for at least one ( c ) between ( -3 ) and ( 4 )

Explanation:

The Intermediate Value Theorem (IVT) states that if a function \( h \) is continuous on a closed interval \([a, b]\), and \( k \) is a number between \( h(a) \) and \( h(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( h(c)=k \).

• For the given function \( h \), it is continuous on \([-3, 4]\), with \( h(-3) = -1 \) and \( h(4) = 2 \).
• We check the value \( k = 1 \): since \( -1 < 1 < 2 \), by IVT, there must be at least one \( c \) in \((-3, 4)\) (the interval of \( x \)-values, not \( y \)-values) such that \( h(c) = 1 \).

Now, analyze the options:

• Option A: Incorrect. IVT refers to \( c \) (the input) being in \([-3, 4]\), not between \( -1 \) and \( 2 \) (which are output values).
• Option B: Incorrect. \( -2 \) is not between \( h(-3) = -1 \) and \( h(4) = 2 \), so IVT does not guarantee \( h(c) = -2 \).
• Option C: Incorrect. Similar to A, \( c \) should be in \([-3, 4]\), and \( -2 \) is not between \( -1 \) and \( 2 \).
• Option D: Correct. \( 1 \) is between \( -1 \) and \( 2 \), so IVT guarantees \( h(c) = 1 \) for some \( c \) in \([-3, 4]\).

Answer:

D. \( h(c) = 1 \) for at least one \( c \) between \( -3 \) and \( 4 \)