1.3 Class Notes

# 1.3 Class Notes - HWK#2 Due Wed 2/7 Thurs 2/8 4.1 14 18 31...

• Notes
• 3

This preview shows pages 1–3. Sign up to view the full content.

HWK #2. Due Wed 2/7, Thurs 2/8 4.1: 14, 18, 31, 71 4.2: 8, 12, 20, 28, 31 HWK #3. Due Tues 2/13, Wed 2/14 Solutions will be available 2/15 4.3: 22, 28, 36, 46 4.4: 2, 8, 18, 38b Not HWK. 4.5: 1, 7, 9, 11 Prelim 1: Tues 2/20 covers 3.3-4.5 4.1. Computing Derivatives, part 2 Rule 1: (x p ) = px p-1 p.214 Rule 2: (cf) = cf p.216 Rule 3. (f+g) = f + g p.217 Cost functions Suppose the cost to produce x units of a particular item is C(x) = 9 + .3 x + .01 x 2 C(x+1)-C(x) = the actual cost of producing one more item C (x) = .3 + .02x, the marginal cost Cost C(x) = 9 + .3 x + .01 x 2 marginal cost C (x) = .3 + .02x 9 is a fixed cost that must be paid to produce any items. C (0) = 0.3. The first item costs 30 cents to make. C (20) = 0.7 The 21 st item costs 70 cents to make Tangent line The equation of the tangent line to the graph at the point x 0 is (p.189) y = f(x 0 ) + f (x 0 ) (x - x 0 ) Why? The value at x = x 0 is f(x 0 ) y = [f(x 0 ) - f (x 0 ) x 0 ] + f (x 0 ) x So the slope is f (x 0 ) If y = a + bx : a = intercept, b = slope Two equations for a line y = a + bx Traditional form of line b= slope, a = intercept y = f(x 0 ) + f (x 0 ) (x - x 0 ) Easier to calculate f (x 0 ) = slope, y = f(x 0 ) at x 0

This preview has intentionally blurred sections. Sign up to view the full version.

y = f(x 0 ) + f (x 0 ) (x - x 0 ) C(x) = 9 + .3 x + .01 x 2 C (x) = .3 + .02x C(20) = 9 + .3(20) + .01(400) = 19 C (20) = .3 + .02(20) = .7
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern