Help in College Level Calculus
[From: ] [author: ] [Date: 11-04-30] [Hit: ]
Determine the function values at the endpoints of the interval 3. Compare the values found in steps 1 and 2. The smallest of all the values is the min for the interval, the largest is the max for the interval.----Well, if it asks for the absolute values then you have to find them.......
f(-2) = -14
f(1) = -5
These are the values you WOULD compare to the values at the critical points, if this function had any. Since it doesn't, these 2 represent the max/min on the interval.
The second problem is going to have critical points. Without doing the work for you I can outline how to solve it.
1. Find all critical points, determine the function values at these points.
2. Determine the function values at the endpoints of the interval
3. Compare the values found in steps 1 and 2. The smallest of all the values is the min for the interval, the largest is the max for the interval.
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Well, if it asks for the absolute values then you have to find them. The process is the same whether looking for an absolute or relative min/max.
You determine whether you have absolute min/max by comparing all the min/max values. You determine local min/max through use of the derivative.
Remember, a minimum, maximum or inflection point will occur where the first derivative is zero OR at the ends of the domain.
a:
f(x) = 3x-8 is a straight line. So it will have a min/max at the ends of the interval . The minimum value is at x = -2 and the maximum is at x = 1. In other words, when you say there is no visible min/max you just didn't realize that you werel ooking at it.
b:
f(x)= x^3-4x^2+5x+9; [-2,3]
f'(x) = 3x^2 -8x + 5. The zeros of f'(x) are at x=1 and x = 5/3.
f''(x) = 6x - 8.
Check the ends of the interval and the zeros of f'(x) to see which is a min/max.
f(-2) = -25
f(1) = 11
f(5/3) = 293/27 ≅ 10.85
f(3) = 15
Obviously the smallest of these is f(-2) and the largest if f(3), both of which occur at the ends of the interval.
So, the absolute min is at (-2,25), the absolute max is at (3,15).
Checking the second derivative, or just looking at the graph of the function confirms that (1,11) is a relative min and (5/3, 293/27) is a relative max.
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