#Candidates Test for Global Extrema

#Table of Contents

1. Introduction
2. Key Concepts
3. Steps for the Candidates Test
4. Exam Tip
5. Worked Example
6. Practice Questions
7. Glossary
8. Conclusion and Key Takeaways

#Introduction

The candidates test is a method used to determine the global extrema (maximum and minimum) of a continuous function on a closed interval. This is based on the extreme value theorem, which guarantees the existence of at least one global maximum and one global minimum for continuous functions on closed intervals.

#Key Concepts

#Steps for the Candidates Test

1. Check Continuity: Ensure that the function $$f(x)$$ is continuous on the interval $$[a, b]$$.
2. Find Critical Points: Determine the critical points of $$f(x)$$ on the interval $$(a, b)$$ by solving $$f'(x) = 0$$.
3. Evaluate Function Values: Calculate the values of $$f(x)$$ at the critical points and at the endpoints $$x = a$$ and $$x = b$$.
4. Determine Global Extrema: Compare these values:
   • The largest value is the global maximum.
   • The smallest value is the global minimum.

#Worked Example

Use the candidates test to find the global minimum and global maximum on the graph of the function $$f$$ defined by: $$f(x) = -x^3 + 2x^2 + 4x - 8 \quad \text{for} \quad -3 \leq x \leq 3$$

#Steps

1. Check Continuity:

   • The function $$f(x)$$ is a polynomial, which is continuous on the interval $$[-3, 3]$$.

2. Find Critical Points:

   • Find the first derivative: $$f'(x) = -3x^2 + 4x + 4$$
   • Set the derivative to zero: $$-3x^2 + 4x + 4 = 0$$
   • Solve for $$x$$: $$x = 2$$ and $$x = -\frac{2}{3}$$

3. Evaluate Function Values:

   • At critical points:
     • $$f(2) = -2^3 + 2(2)^2 + 4(2) - 8 = 0$$
     • $$f\left(-\frac{2}{3}\right) = -\left(-\frac{2}{3}\right)^3 + 2\left(-\frac{2}{3}\right)^2 + 4\left(-\frac{2}{3}\right) - 8 = -\frac{256}{27} \approx -9.481$$
   • At endpoints:
     • $$f(-3) = -(-3)^3 + 2(-3)^2 + 4(-3) - 8 = 25$$
     • $$f(3) = -3^3 + 2(3)^2 + 4(3) - 8 = -5$$

4. Determine Global Extrema:

   • Global maximum: $$25$$ at $$x = -3$$
   • Global minimum: $$-\frac{256}{27} \approx -9.481$$ at $$x = -\frac{2}{3}$$

- **Global Maximum**: $$(-3, 25)$$ - **Global Minimum**: $$\left(-\frac{2}{3}, -\frac{256}{27}\right)$$

#Practice Questions

Practice Question

1. Find the global extrema of the function $$g(x) = x^3 - 6x^2 + 9x + 1$$ on the interval $$[0, 4]$$.

Practice Question

2. Determine the global maximum and minimum of $$h(x) = 2x^2 - 4x + 1$$ on $$[-2, 3]$$.

#Glossary

• Continuous Function: A function with no breaks, jumps, or holes in its domain.
• Critical Point: A point where the first derivative of the function is zero or undefined.
• Endpoint: The boundary points of a closed interval.
• Global Extrema: The highest or lowest value of a function over a given interval.
• Extreme Value Theorem: A theorem that guarantees the existence of at least one global maximum and one global minimum for continuous functions on closed intervals.

#Conclusion and Key Takeaways

• The candidates test is a reliable method for finding global extrema by evaluating the function at critical points and endpoints.
• Always ensure the function is continuous on the interval before applying the candidates test.
• Critical points and endpoints are both essential in determining the global extrema.

By following these steps and using the candidates test, you can confidently determine the global extrema for any continuous function on a closed interval.