How do you do this?
By definition, F'(x) is the limit as h approaches zero of (F(x+h)-F(x))/h. We can approximate this limit with the difference quotient -- e.g. one can take h = .2 and estimate F'(1.2) as (F(1.2 + .2)-F(1.2))/.2 = (.1657-.09098)/.2 = .37360. The remaining values of F'(x) can be estimated in the same manner.
Please explain.
The derivative of F(x) is f(x). So, f(x) > 0 and decreasing on 1 < x < 3 means that F'(x) > 0 and decreasing on 1 < x < 3. Since the derivative is positive on the interval, the function F(x) is increasing. Since the derivative is decreasing on 1 < x < 3, the function F(x) is concave down on that interval.
Similarly, since F'(x) is negative and decreasing on 3 < x < 7, we know that F(x) is decreasing and concave down on that same interval. Because F(x) is increasing to the left of 3 and decreasing to the right of 3 and because F(x) is continuous at 3 (because F(x) is differentiable there), we know by the first derivative test that F has a relative maximum there.
How do you do this problem?
The rate of sales is the function p(t) where p(t) = 1000 for 8 <= t < 10, p(t) = 1500 for 10 <= t < 12, p(t) = 0 for 12 <= t < 13, p(t) = 900 for 13 <= t < 15, and p(t) = 500 for 15 <= t < 17 (where I am expressing time in hours since midnight.) The job is to estimate the integral of p(t) between 8 and 17 (The text of the problem says 0 to 9 because he is measuring time in hours from 8:00 A.M.)
Using the left rectangular sum, with subdivision points at 8, 10, 12, 1, 13, 15, and 17 gives the estimate 2*1000 + 2*1500 + 1*0 + 2*900 + 2*500 = 7800 dollars. The average total sales is the integral divided by 8, because you are not counting lunch. So the average would be 7800/8 = 975 dollars per hour.
How do you do this problem?
Again, the idea is to divide up the interval from 0 to 8 weeks, which is 0 to 8/52 years by subdividing at t = 0, t = 2/52, t = 6/52, and t = 8/52. The estimate is (2/52)*50 + (4/52)*30 + (2/52)*10 plus 6.5.
Please explain the tangent line approximation.
The idea is: Near x = a, the graph of the tangent line to y = f(x) at x = a is a good approximation of the graph of the function y = f(x).
For example, suppose we want to approximate y = e^x near x = 0.5. Then The tangent line at x = 0.5 is the line with slope equal to the derivative of the function at 0.5 and which passes through the point (.5,e^(.5)). So, the line is y - e^(.5) = e^(.5)(x - .5). So e^x for x near 0.5 would be approximately equal to e^(0.5) + e^(0.5)(x - 0.5). If we wanted to estimate e^(0.75) with this formula, then we could put x = 0.75. But the problem is that we would still need to estimate e^(0.5). Luckily, the problem has just calculated an estimate for this, viz. e^(0.5) is approximately 1.5625. Using this, we get an approximation 1.5625+1.5625(0.75 - 0.5)=1.953125. One can now repeat the process to approximate the value of e^1 using the our approximation to e^0.75 and the tangent line approximation to e^x at x = 0.75.