The First Derivative Test
The First Derivative Test (Motivation and Theorem) If f is a function, then f has a relative maximum at x = c if for all points a near c, f(c) > f(a), and f has a relative minimum at x = c if for all points a near c, f(c) < f(a). Consider a relative maximum, we have that on the left, the function is increasing and on the right the function is decreasing. Similarly, for a relative minimum, on the right the function is decreasing and on the left the function is increasing.
We can now state the first derivative test:
Example
Solution
So there is a critical point at 5/7. Notice
also that there is a critical point at x = 1 since the first
derivative is undefined there (notice the negative exponent 3/5). To determine whether the critical
point is a relative max, min or neither, choose a number just above and just
below the critical value.
The actual graph is shown below
Exercise Classify the relative extrema of f(x)= x + 1/x
Answer (Hold your mouse on the yellow rectangle)
Global Extrema
To determine the global maximum and minimum we proceed as follows: 1) Find all critical points of f and write down the y values in a table 2) Find f(a) and f(b) and add them to the table. 3) The largest number in the table will be the global maximum, and the smallest number in the table will be the global minimum.
Example Find the Global Max and Min of f(x) = 2x^{3} +3x^{2}  36x + 2 on the interval [10,3]
Solution We compute f '(x) = 6x^{2} + 6x  36 = 6(x  2)(x + 3) Hence there are critical points at 2 and 3. We compute f(2) = 42, f(3) = 83, f(10) = 1942, f(3) = 25
Hence on [10,3], f has a global minimum of 1338 at x = 10 and a global maximum of 83 at x = 3. Below is the graph.
Exercise: Find the absolute extrema of x^{3}  12x on the interval [0,4].
Answer (Hold your mouse on the yellow rectangle)
Application The cost in cents of supplying x burgers is C = .5x^{2} + 100 0 < x < 200 Determine the number of burgers that will minimize the average cost.
Solution The average cost is given by the Cost divided by the number of units. We have Ave = C/x = 0.5 x + 200/x Take a derivative to find the critical points Ave ' = 0.5  200/x^{2} = 0 Multiply by x^{2} 0.5 x^{2}  200 = 0 x^{2} + 400 = 0 x = 20 Now use a table to determine whether the minimum occurs at the critical point or the endpoint.
We can see that supplying 20 burgers will give the minimum average cost of 20 cents per burger.
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