Limits at Infinity and Asymptotes
- x=1x=1
- x=−1,x=2
- x=0
- Yes, there is a vertical asymptote
- Yes, there is a vertical asymptote
- 0
- ∞
- −17
- –2
- –4
- Horizontal: none, vertical: x=0
- Horizontal: none, vertical: x=±2
- Horizontal: none, vertical: none
- Horizontal: y=0, vertical: x=±1
- Horizontal: y=0, vertical: x=0 and x=−1
- Horizontal: y=1, vertical: x=1
- Horizontal: none, vertical: none
- Answers will vary, for example: y=2xx−1
- Answers will vary, for example: y=4xx+1
- y=0
- ∞
- y=3
Applied Optimization Problems
- The critical points can be the minima, maxima, or neither.
- False; y=−x2 has a minimum only
- h=623 in.
- 1
- 100 ft by 100 ft
- 40 ft by 40 ft
- 19.73 ft.
- T(θ)=40θ3v+40cosθv
- approximately 34.02 mph
- 4
- 0
- Maximal: x=5,y=5; minimal: x=0,y=10 and y=0,x=10
- Maximal: x=1,y=9; minimal: none
- 4π3√3
- 6
- r=2,h=4
- (2,1)
- (0.8351,0.6974)
- A=20r−2r2−12πr2
- C(x)=5x2+32x
- P(x)=(50−x)(800+25x−50)
L’Hôpital’s Rule
- ∞
- 12a
- 1nan−1
- Cannot apply directly; use logarithms
- Cannot apply directly; rewrite as limx→0x3
- 6
- –2
- –1
- n
- −12
- 12
- 1
- 16
- 1
- 0
- 0
- –1
- ∞
- 0
- 1e
- 0
- 1
- 0
- tan(1)
- 2
Newton’s Method
-
- F(xn)=xn−(xn)3+2xn+13(xn)2+2
- F(xn)=xn−exnexn
- |c|>0.5 fails, |c|≤0.5 works
- c=1f′(xn)
-
- x1=1225,x2=312625
- x1=−4,x2=−40
-
- x1=1.291,x2=0.8801
- x1=0.7071,x2=1.189
-
- x1=−2625,x2=−1224625
- x1=4,x2=18
-
- x1=610,x2=610
- x1=2,x2=2
- 3.1623 or –3.1623
- 0, –1, or 1
- 0
- 0.5188 or –1.2906
- 0
- 4.493
- 0.159, 3.146
- To find candidates for maxima and minima, we need to find the critical points f′(x)=0. Show that to solve for the critical points of a function f(x), Newton’s method is given by xn+1=xn−f′(xn)f′′(xn).
- We need f to be twice continuously differentiable.
- x=0
- x=−1
<lix=5.619
- x=−1.326
- There is no solution to the equation.
- It enters a cycle.
- 0
- –0.3513
- Newton: 11 iterations, secant: 16 iterations
- Newton: three iterations, secant: six iterations
- Newton: five iterations, secant: eight iterations
- E=4.071
- 4.394%
Antiderivatives
- F′(x)=15x2+4x+3
- F′(x)=2xex+x2ex
- F′(x)=ex
- F(x)=ex−x3−cos(x)+C
- F(x)=x22−x−2cos(2x)+C
- F(x)=12x2+4x3+C
- F(x)=25(√x)5+C
- F(x)=32x2/3+C
- F(x)=x+tan(x)+C
- F(x)=13sin3(x)+C
- F(x)=−12cot(x)−1x+C
- F(x)=−secx−4cscx+C
- F(x)=−18e−4x−cosx+C
- −cosx+C
- 3x−2x+C
- 83x3/2+45x5/4+C
- 14x−2x−12x2+C
- f(x)=−12x2+32
- f(x)=sinx+tanx+1
- f(x)=−16x3−2x+136
- Answers may vary; one possible answer is f(x)=e−x
- Answers may vary; one possible answer is f(x)=−sinx
- 5.867 sec
- 7.333 sec
- 13.75 ft/sec2
- F(x)=13x3+2x
- F(x)=x2−cosx+1
- F(x)=−1x+1+1
- True
- False