Example: Let g(x) = x 3 + 2. a) Find g'(x) for any x value. b) - - PDF document

‡ Example: Let g(x) = x3 +2. a) Find g'(x) for any x value. b) Find g'(2) = slope of the tangent line to the graph y=g(x) at the point (2,g(2)). c) Determine the equation of the line tangent to the graph of y=g(x) at the point (2,g(2)). ‡ Solutions: a) Calculate g'(x) in stages. First get the difference quotient over the interval [x,x+h]:

gHx+ hL- gHxL

Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä

h

=

AHx+ hL3+ 2E -@x3+ 2D

Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä

h

= x 3+

3 x2 h+ 3 xh2+ h3+ 2- x 3- 2

Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä

h

= 3

x2 h+ 3 xh2+ h3

Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä

h

=3 x2 +3 xh +h2. The derivative is the limit of this difference quotient as hÆ0: g'(x) = limhÆ0(3 x2 +3 xh +h2)=3x2. ‡ b) To find the slope at the point where x=2, simply compute g'(2) = 3H2L2 =12. This says that the tangent line at (2,g(2)) has slope = 12. c) Since g(2) = H2L3 +2 =10, the point at which we wish to build the tangent line is (2,g(2)) = (2,10). At this point the slope will be g'(2)=12 from b).

temp.nb 1

Use point slope formula: Hy -y1L =mHx -x1L which is y-10 = 12(x-2) or y=12x-14. (equation of tangent line) Example : The total cost, C HxL Hin dollarsL incurred by the Aloha Company in manufacturing x surfboards per day is given by C HxL =- 10 x3 + 300 x + 130 H0 £x £15L.

temp.nb 2

‡ a) Find C'(x) b) What is the rate of change of the total cost when the level of production is 10 surfboards per day? c) What is the average per surfboard cost which Aloha incurs in manufacturing 10 surfboards per day? ‡ a) Again with the difference quotient:

CHx+ hL- CHxL

Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä

h

=

A- 10 Hx+ hL2+ 300 Hx+ hL+ 130E- @- 10 x 2+ 300 x+ 130D

Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä

h

= -

10 x 2- 20 xh- 10 h2+ 300 x+ 300 h+ 130+ 10 x2- 300 x- 130

Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä

h

= -

20 xh- 10 h2+ 300 h

Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä

h

= -20x-10h+300 Now our derivative, C'(x) is the limit of this difference quotient as hÆ0. C'(x) = limhÆ0(-20x-10h+300) = -20x+300. b) Since C'(x) provides us with the rate at which C changes per surfboard when we make x surfboards per week, we seek C'(10), the rate at which C changes when we make 10 per week. C'(10) = -20(10)+300 = 100 dollars per surfboard. We can interpret this to mean that when we make the eleventh board, our total costs will increase by $100. ‡ c) The average cost of making 10 surfboards will be the amount by which our costs increase in going from 0 to 10 surfboards, divided by the number of surfboards (10). average cost of 10 boards = CH10L-

CH0L

Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä

10- temp.nb 3

= -

10 H10L2+ 300 H10L+ 130- 0- 0- 130

Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä

10

= $200 per surfboard.

temp.nb 4