Question

evaluate (f(x) = x^3 - 5x^2 + 3x + 6) at -1 and -3 to determine if the intermediate value theorem guarantees that a zero exists between the two values.

a) (f(-1) =)

b) (f(-3) =)

c) does the intermediate value theorem guarantee that a zero exists between -1 and -3?

yes
no

Explanation:

Step1: Evaluate \(f(-1)\)

$$f(-1) = (-1)^3 - 5(-1)^2 + 3(-1) + 6 = -1 - 5 - 3 + 6 = -3$$

Step2: Evaluate \(f(-3)\)

$$f(-3) = (-3)^3 - 5(-3)^2 + 3(-3) + 6 = -27 - 45 - 9 + 6 = -75$$

Step3: Apply Intermediate Value Theorem

Since \(f(-1) = -3 < 0\) and \(f(-3) = -75 < 0\), both values have the same sign, so the theorem does not guarantee a zero.

Answer:

a) \(f(-1) = -3\)
b) \(f(-3) = -75\)
c) no