Logarithms
Write the expression as a logarithm of a single quantity.
(3/2)[ln(x2 + 1) - ln(x + 1) - ln(x - 1)]
ln[((x2 + 1) / (x2 - 1))3]1/2
Find the general solution of the differential equation.
dy/dx = xe9x^2
y = (1/18)e9x^2 + C
Find the derivative of the function.
h(t) = log5(4 - t)2
h'(t) = 2 / [(t - 4)ln5]
Evaluate the expression WITHOUT using a calculator.
arctan(31/2 / 3)
π/6
After t years, the value of a car purchased for $25,000 is V(t) = 25000(3/4)t.
a. Determine the value of the car 2 years after it was purchased.
b. Find the rates of change of V with respect to t when t = 1 and t = 4.
c. Determine the horizontal asymptote of V'(t). Interpret its meaning in the context of the problem.
a. $14,062.50
b. When t = 1, dV/dt = -5394.04.
When t = 4, dV/dt = 2275.61.
c. Horizonal Asymptote: V' = 0
As the car ages, it is worth less each year and depreciates less each year, but the value of the car will never reach $0.
Find the derivative of the function.
f(x) = ln|sin(x)|
dy/dx = cot(x)
Find the indefinite integral.
∫((e-x) / (1 + e-x)) dx
x - ln(ex + 1) + C
Find the derivative of the function.
g(x) = log5[4 / (x2(1 - x)1/2)]
g'(x) = (-2 / (xln5)) + (1 / (2(1 - x)ln5)
Find any relative extrema of the function.
arcsin(x) - 2x
Relative Minimum: (31/2/2, -0.68)
Relative Maximum: (-(31/2/2), 0.68)
The velocity of v of an object falling through a resisting medium such as air or water is given by
v = (32/k)[1 - e-kt+ ((v0ke-kt) / 32)]
where v0 is the initial velocity, t is the time in seconds, and k is the resistance constant of the medium. Find the formula for the velocity of a falling body in a vacuum by fixing v0 and t and letting k approach 0.
Hint - use L'Hopital's Rule
limk → 0 = 32t + v0
Find an equation of the tangent line to the graph of the function at the given point.
f(x) = ln[(1 + sin2(x))1/2], (π/4, ln[(3/2)1/2])
y - ln[(3/2)1/2] = (1/3)(x - (π/4))
or
y = (x/3) + (1/2)ln(3/2) - (π/12)
exy = x2 - y2 = 10
dy/dx = -[(yexy + 2x) / (xexy - 2y)]
Find dy/dx.
y = (1 + x)1/x
[(1 + x)1/x / x][(1 / (x + 1)) - (ln(x + 1) / x)]
Find the indefinite integral.
∫[(x + 5) / ((9 - (x -3)2)1/2)] dx
-(6x - x2)1/2 + 8arcsin((x/3) - 1) + C
Find the value of a such that the area bounded by y = e-x, the x-axis, x = -a, and x = a is 8/3.
a = ln3
Determine whether the statement is true or false. if it is false, explain why or give an example that shows it is false.
(d/dx)[ln(cx)] = (d/dx)[lnx], where c > 0.
True. ln(cx) = lnc + lnx and
(d/dx)[ln(cx)] = 1/x = (d/dx)[lnx].
Find the relative extrema and the points of inflection (if any exist) of the function.
g(x) = (1/(2π)1/2)e-[((x - 3)^2) / 2]
Relative Maximum: (3, (1/(2π)1/2)
Points of Inflection: (2, (1/(2π)1/2)e-1/2) and
(4, (1/(2π)1/2)e-1/2)
Find the indefinite integral.
∫[32x / (1 + 32x)] dx
(1 / 2ln3)[ln(1 + 32x)] + C
Find the area of the given region.
y = (3cos(x)) / (1 + sin2(x)), -π/2 < x < π/2
3π/2
A billboard 85 feet wide is perpendicular to a straight road and is 40 feet from the road. Find the point on the road at which the angle θ subtended by the billboard is a maximum.
There is a maximum of x = 50(21/2) = 70.71 ft
The relationship between the number of decibels b and the intensity of a sound I in watts per centimeter squared is
b = (10/ln10)[ln(I / (10-16))]
Determine the number of decibels of a sound with an intensity of 10-5 watt per square centimeter.
110 decibels
Verify that the function
y = L / (1 + ae-x/b), a > 0, b > 0, L > 0
increases at a maximum rate when y = L/2
y' = ((aL/b)e-x/b) / (1 + ae-x/b)2
y'' = 0 if ae-x/b = 1 -----> x = blna
y(blna) = L/2
Therefore, the y-coordinate of the inflection point is L/2.
Determine y' given yx = xy.
y' = (y2 - xy lny) / (x2 - xy lnx)
A region's area is represented by
∫01 arcsinx dx.
Find the exact area analytically.
(π/2) - 1
Let f(x) = [(ax - 1) / (ax + 1)] for a > 0, a ≠ 1.
Show that f has an inverse function. Then fine f-1.
y' = f'(x) = [(2axlna) / (ax + 1)2]
For 0 < a < 1, y' < 0 -> one-to-one has an inverse.
For a > 1, y' > 0 -> one-to-one has an inverse.
f-1(x) = (1 / lna)[ln((x + 1) / (1 - x))]