Question

given the function: f(x) = x³ - 4x² + 3x + 8 on the closed interval -1, 2, if k = 8 is an intermediate value, find the number c guaranteed by the intermediate value theorem. c = ?

Explanation:

Step1: Set up the equation

We need to find \( c \) in \([-1, 2]\) such that \( f(c)=8 \). So we set up the equation:

$$ c^{3}-4c^{2}+3c + 8=8 $$

Step2: Simplify the equation

Subtract 8 from both sides of the equation:

$$ c^{3}-4c^{2}+3c+8 - 8=8 - 8 $$

$$ c^{3}-4c^{2}+3c=0 $$

Step3: Factor the equation

Factor out a \( c \) from the left - hand side:

$$ c(c^{2}-4c + 3)=0 $$

Then factor the quadratic expression \( c^{2}-4c + 3 \). We need two numbers that multiply to 3 and add up to - 4. The numbers are - 1 and - 3. So:

$$ c(c - 1)(c - 3)=0 $$

Step4: Solve for c

Using the zero - product property, if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \). So we have three solutions for \( c \):
\( c=0 \), \( c - 1=0\Rightarrow c = 1 \), and \( c - 3=0\Rightarrow c = 3 \)

Step5: Check the solutions in the interval

We are looking for \( c \) in the interval \([-1,2]\).

• For \( c = 0 \), \( 0\in[-1,2] \)
• For \( c = 1 \), \( 1\in[-1,2] \)
• For \( c = 3 \), \( 3

otin[-1,2] \) (since \( 3>2 \))

Answer:

\( c = 0 \) or \( c = 1 \) (both \( 0 \) and \( 1 \) are in the interval \([-1,2]\) and satisfy \( f(c)=8\))