Extreme Value Theorem - Intermediate Value Theorem

Basic calculus

1. Intermediate Value Theorem

2. Example 1: Is there a solution to between and ?   First, check if the given is continuous. Evaluate the functions using the given endpoints, then check if zero is in between the two computed values.     because , then, according to IVT, the function has at least one solution between and .  

3. Example 3: Does the function have a solution on the closed interval [0, 4]?   First, check if the given is continuous. Evaluate the functions using the given endpoints, then check if one is in between the two computed values.     Since 1 is not between (0) and (4), the IVT fails to hold for the given function on the closed interval [0, 4] and with  

4. Example 3: Does the function have a solution on the closed interval [-2, 2]?   First, check if the given is continuous. Evaluate the functions using the given endpoints, then check if one is in between the two computed values.     Since , that is , then yes, the function has a solution on [-2, 2].  

5. Extreme Value Theorem

6. Extreme Value Theorem If is continuous at every point of a closed interval [a, b], then assumes both an absolute maximum value M and an absolute minimum value m somewhere in [a, b]. That is, there are numbers 𝑐1 and 𝑐2 within [a, b], such that and , and for every other x in the interval. The highest value of a function f on a given closed interval is called its absolute maximum value, and its lowest value is called its absolute minimum value. Collectively, these values are known as extreme values of on a closed interval  

7. Identify the extreme values of the function on the closed interval [-2, 1] shown in Figure 2. Solution: On the given interval, the graph’s lowest value is 1 which occurs on two values of and its highest value is 4 which occurs when . Thus, the absolute minimum value of the function is 1 and the absolute maximum value is 4.   Example #1

8. Identify the absolute extreme values of the function defined on the interval [-3, 2]. Solution: First, sketch the graph of the function on the given interval, similar to the one in Figure 3. Refer to the table of values below.   Example #2 Based on the figure, the function f has an absolute maximum value of 0 and an absolute minimum value of -9 on the interval [-3, 2].

9. Example 6: At a school canteen, bottled water is priced P20.00 for students. At that price, the canteen sells 200 items daily. For every P2.00 increase in price, there are 10 fewer students willing to buy the bottled water. With this scheme, what will be the revenue function of the school canteen for the sale of bottled water? Solution: a. The revenue function is simply the product of the selling price and the quantity of items sold . Without the P2.00 increase, the revenue of the canteen is fixed at P20.00 X 200 = P4,000.00 daily. Let x represent the number of times that the canteen increases the price of bottled water. 𝑅(𝑥) = 𝑠𝑒𝑙𝑙𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒 𝑋 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑡𝑒𝑚𝑠 𝑠𝑜𝑙𝑑  

10. Example 6: At a school canteen, bottled water is priced P20.00 for students. At that price, the canteen sells 200 items daily. For every P2.00 increase in price, there are 10 fewer students willing to buy the bottled water. b. You are asked to determine the absolute maximum value of the function on the interval [0, 10]. To answer the question, you can set up a table of values for the function and sketch the graph.   If you observe the table above, the maximum revenue of the canteen will happen after a P2.00 increase is done 5 times; thus, it will reach a total price of P30.00 If the canteen decides to increase the price 10 times, the revenue will be the same as that when sold by the original price.

11. The Equation of the Tangent Line The precise definition of a tangent line relies on the notion of a secant line. Let C be the graph of a continuous function and let P be a point on C. A secant line to through P is any line connecting P and another point Q on C. In the figure on the right, the line P Q is a secant line of through P.  

12. The Equation of the Tangent Line If the sequence of secant lines to the graph of through P approaches one limiting position (in consideration of points Q to the left and from the right of P), then we define this line to be the tangent line to at P.  

13. The Equation of the Tangent Line Consider the graph of a function whose graph is given below. Let be a point on the graph of . Our objective is to find the equation of the tangent line (TL) to the graph at the point  

14. The Equation of the Tangent Line Since the tangent line is the limiting position of the secant lines as Q approaches P, it follows that the slope of the tangent line (TL) at the point P is the limit of the slopes of the secant lines as x approaches . In symbols Finally, since the tangent line passes through then its equation is given by  

15. Example 1: Find the equation of the tangent line to          

16. Example 2: Find the equation of the tangent line to the graph of