Math Problem Statement

Decide whether the Intermediate Value Theorem applies to the given function on the interval [0, 6]: f(x) = -x^2 + x + 22, f(c) = 2.

Solution

The question asks whether the Intermediate Value Theorem (IVT) applies to the given function f(x)=x2+x+22f(x) = -x^2 + x + 22 on the interval [0,6][0, 6], and if so, to find cc such that f(c)=2f(c) = 2. The possible answers involve whether the theorem applies and the value of cc.

To apply the IVT:

  1. Conditions for IVT:
    • The function must be continuous on the closed interval [0,6][0, 6].
    • If the function is continuous, the IVT states that for any value between f(0)f(0) and f(6)f(6), there exists some c[0,6]c \in [0, 6] such that f(c)f(c) equals that value.

Step-by-step verification:

  1. Check continuity: The function f(x)=x2+x+22f(x) = -x^2 + x + 22 is a quadratic function, which is continuous everywhere, so it is continuous on [0,6][0, 6].

  2. Evaluate f(0)f(0) and f(6)f(6):

    • f(0)=(0)2+(0)+22=22f(0) = -(0)^2 + (0) + 22 = 22
    • f(6)=(6)2+6+22=36+6+22=8f(6) = -(6)^2 + 6 + 22 = -36 + 6 + 22 = -8
  3. Check if 2 is between f(0)f(0) and f(6)f(6):

    • f(0)=22f(0) = 22 and f(6)=8f(6) = -8. Since 22 lies between 8-8 and 2222, the IVT guarantees that there is some c[0,6]c \in [0, 6] such that f(c)=2f(c) = 2.

Find cc such that f(c)=2f(c) = 2:

Solve x2+x+22=2-x^2 + x + 22 = 2: x2+x+22=2-x^2 + x + 22 = 2 x2+x+222=0-x^2 + x + 22 - 2 = 0 x2+x+20=0-x^2 + x + 20 = 0 Multiply the equation by 1-1 to simplify: x2x20=0x^2 - x - 20 = 0 Now solve the quadratic equation using the quadratic formula: x=(1)±(1)24(1)(20)2(1)x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-20)}}{2(1)} x=1±1+802x = \frac{1 \pm \sqrt{1 + 80}}{2} x=1±812x = \frac{1 \pm \sqrt{81}}{2} x=1±92x = \frac{1 \pm 9}{2} Thus, x=1+92=5x = \frac{1 + 9}{2} = 5 or x=192=4x = \frac{1 - 9}{2} = -4.

Since cc must be in the interval [0,6][0, 6], the solution is c=5c = 5.

Conclusion:

The Intermediate Value Theorem applies, and the value of cc is 55, matching the third option.


Do you have any questions or would you like more details?

Follow-up Questions:

  1. How does the Intermediate Value Theorem work for more complex functions?
  2. Why is continuity a critical requirement for the IVT?
  3. Can the IVT be applied if the function is discontinuous?
  4. How would you solve this problem graphically?
  5. What other mathematical theorems are closely related to the IVT?

Tip: When solving quadratic equations, always check the discriminant to see if real solutions exist before applying the quadratic formula!

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Math Problem Analysis

Mathematical Concepts

Calculus
Intermediate Value Theorem
Continuity

Formulas

Quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a

Theorems

Intermediate Value Theorem

Suitable Grade Level

College Level (Calculus I)