Ch 1
The limit of sqrt(x) as x-->9
3
The derivative of x2 - 2x + 3
2x - 2
Find the absolute extrema of the following function the closed interval:
g(x) = 2x + 5cos(x) [0, 2pi]
Maximum: (2pi, 17.57)
Minimum: (2.73, 0.88)
For the function F'(x), find the original function F when
F'(x) = (-1/2sin(x/2))sin(cos(x/2))
F(x) = -cos(cos(x/2)) + C
Solve the differential equation fro y.
y' = (x^(5/2))(7y)
y = sqrt((4/49)x^(7/2) + C)
The limit of (x3 - 12) as x-->5
113
The derivative of x1/2 - x-1/2
(1/2)x(-1/2) + (1/2)x(-3/2)
Find the points guaranteed by the Mean Value Theorem:
f(x) = x - cos(x) [-pi/2, pi/2]
f'(0) = 1
The number of toys produced by a factory over the interval (0,8) is given by the function A(t) = 15x3-3x2+cos(2x), where t is given in hours. Find the average number of toys produced by the factory, per hour, over this 8-hour interval.
1855 toys
Solve the differential equation for y.
dy/dx = (7 - y)x
The limit of ((x2 - 2x)/x) as x-->0
-2
f'(a) when f(a) = 3cos(a) - (sin(a)/4)
-3sin(a) - (cos(a)/4)
Find the relative extrema of the function:
h(t) = (1/4)t4 - 8t
Relative Minimum at (2,-12)
Let g(x) be the derivative of f(x). If f(x) equals the integral of 3h(t) from 5 to g(x), find g(x)
g(x) = 3h(g(x))g'(x)
Use integration to find a general solution to the differential equation
y' = (1/3)x5tan(x6)
y = (1/18)ln|sec(x6)| + C
The limit of (t + 2)/(t2 - 4) as t-->-2
-1/4
To estimate the height of the building, a weight is dropped from the top of the building into a pool at ground level. How high is the building if the splash is seen 9.2 seconds after the weight is dropped?
414.74 m or 1354 ft
Determine the points of inflection and discuss concavity of the function:
f(x) = x + cos(x)
POIs: (pi/2, pi/2); (3pi/2, 3pi/2)
Concave up: (pi/2, 3pi/2)
Concave down: (0, pi/2); (3pi/2, 2pi)
2The rate at which leaves fall out of a tree during fall is modeled by the functions F(t) = log7(sec(3t2)). Find A(t), the derivative of F(t)
A(t) = (ln|sec(3t2)+tan(3t2)|) / (ln 7)sec(3t2)
Find the particular solution y = f(x) to the differential equation that satisfies the initial condition.
dy/dx = ey-2x, y(0) = 0
y = ln(2 / (e-2x + 1))
The limit of ([1/(x+1)] - 1) / x as x-->0
-1
Use implicit differentiation to find dy/dx:
xsin(y) = ycos(x)
(ysin(x) + sin(y))/(cos(x) - xcos(y))
The limit of (6x / (x + cosx)) as x-->-infinity
6
Find F(x), the antiderivative of f(x), if f(x) = (x2-2)/(x-1)3 and F(2) = 1
F(x) = ln|x-1| + (1/2)(x-1)-2 + (1/2)
Solve the differential equation for y.
y' = 5x2+3x3-tanx
y = (5/3)x3 + (3/4)x4 + ln|cosx| + C