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Math 105: Review for Final Exam, Part II - SOLUTIONS
- Consider the function f^ f f(((xxx) =) =) = xxx^666 −−− 222 xxx^333 on the interval [[[−−− 222 ,,, 2]2]2].
(a) Find the xxx- and yyy-coordinates of any and all local extrema and classify each as a local maximum or local minimum. f′^ (x) = 6x^5 − 6 x^2 0 = 6x^2 (x^3 − 1) ⇒ x = 0, 1 (and the endpoints x = − 2 , 2 − 2 ≤ x < 0 0 < x < 1 1 < x ≤ 2 f′^ negative negative positive f ↘ ↘ ↗ y-values: f(−2) = 80, f(0) = 0, f(1) = −1, f(2) = 48 So, f has a local minimum at (1, −1) and local maxima at (− 2 , 80) and (2, 48); (0, 0) is not a local extremum.
(b) Find the xxx- and yyy-coordinates of any and all global extrema and classify each as a global maximum or global minimum. We check the y-values at the local extrema and the endpoints. y-values: f(−2) = 80, f(1) = −1, f(2) = 48 So, f has a global minimum at (1, −1) and a global maximum at (− 2 , 80).
(c) Find the xxx-coordinate(s) of any and all inflection points. f′′(x) = 30x^4 − 12 x 0 = 6x(5x^3 − 2) ⇒ x = 0,
x < 0 0 < x <
- 4 < x f′′^ positive negative positive f concave up concave down concave up So, the x-values of the inflection points of f are x = 0 and x =
- How would your answers to the previous question have changed if the domain of fff were all reals? The answers would be the same except that there would no longer be a global maximum. This is because as x → ±∞, f(x) → ∞, (i.e., the y-values grow without bound).
- Use Newton’s Method to find a root of fff(((xxx) =) =) = xxx^333 −−− xxx + 1+ 1+ 1 correct to three decimal places.
Recall that xn+1 = xn −
f(xn ) f′(xn)
= xn −
x^3 n − xn + 1 3 x^2 n − 1
Using a calculator with initial guess x 0 = −1, we get the following. x 0 = − 1. 5 x 1 = − 1. 34782 ... x 2 = − 1. 32520 ... x 3 = − 1. 32471 ... x 4 = − 1. 32471 ... x 5 = − 1. 32471 ...
So, the root is at approximately x = − 1 .324 (truncated) or x = − 1 .325 (rounded).
- Your company is mass-producing a cylindrical container. The flat portion (top and bot- tom) costs 3 cents per square inch and the curved (lateral) portion costs 5 cents per square inch. If your budget is $9.00 per container, what dimensions will give the largest volume? area of circle = πr πr πr^222 lateral area of cylinder = 222 πrhπrhπrh volume of cylinder = πrπrπr^222 hhh Objective function: volume = V = πr^2 h We need to get this down to a function of just one variable, so we use the
constraint equation : cost = 900 = 3 · 2 · πr^2 + 5 · 2 πrh 900 = 6πr^2 + 10πrh 900 − 6 πr^2 = 10πrh 900 − 6 πr^2 10 πr
= h
Substituting this back into the objective function gives
V = πr^2 h = πr^2 ·
900 − 6 πr^2 10 πr
= r ·
900 − 6 πr^2 10
(900r − 6 πr^3 ). Now that we have V as a function of just one variable, we find its minimum.
V ′(x) =
(900 − 18 πr^2 )
0 =
(900 − 18 πr^2 )
⇒ 18 πr^2 = 900
⇒r^2 =
π
⇒r =
π
Since V ′^ is postive for 0 < r <
π
and negative for
π
< r, we know that r =
π
≈ 3 .989 inches is the radius that gives maximum volume.
And h =
900 − 6 πr^2 10 πr = ...much simplifying... =
π ≈ 4 .787 inches.
- You are watching a plane flying toward your position at a constant height of 3 miles and a speed of 500 miles per hour relative to the ground. At the moment when the plane is 5 miles from you (diagonally), at what rate is the angle of your vision toward the plane changing?
a
c 3 θ
Plane
You
We know
da dt
, and we want to find
dθ dt
So, we write an equation that relates a and θ and then differentiate implicitly with respect to time t.
tan θ =
a sec^2 θ
dθ dt
a^2
da dt dθ dt
a^2
da dt
cos^2 θ
time (min) 0 2 4 6 8 rate (gal/min) 15 11 8 4 3
overestimate = L 4 = (15 + 11 + 8 + 4)(2) = 76 underestimate = R 4 = (11 + 8 + 4 + 3)(2) = 52
- The rate of change of a room’s temperature is rrr(((ttt) =) =) = ttt^222 −−− 999 degrees per hour on the interval [0[0[0,,, 4]4]4] hours. At ttt = 0= 0= 0, the temperature is 70 degrees. (Remember that rrr is the derivative of the temperature function.) Let T (t) be the temperature at time t.
(a) When on this interval is the temperature rising? falling?
Note that T ′(t) = r(t) and that r(t) = t^2 − 9 = 0 only when t = 3.
t < 3 3 < t T ′^ negative positive T ↘ ↗ So, we see that the temperature is rising on [0, 3) and falling on (3, 4]. (b) What is the maximum temperature on this interval and when does it occur?
T (t) =
r(t) dt =
t^3 3
− 9 t + C =
t^3 3
− 9 t + 70 C = 70 because we know T (0) = 70.
From our number line in part(a), we see that the maximum must occur at an endpoint. T (0) = 70 and T (4) = 55
, so the maximum temperature of 70 degrees occurs at t = 0.
(c) What is the minimum temperature on this interval and when does it occur? From our number line in part(a), we see that the minimum must occur at t = 3. That temperature is T (3) = 52 degrees. (d) What is the average rate of change of the temperature on this interval? ∫ (^4) 0 r(t)^ dt 4 − 0
T (4) − T (0)
degrees per hour
- Use sigma notation to express LLL 101010 as approximations to
20
ln x dx
20
ln x dx
20
ln x dx.
We’re subdividing the interval into 10 pieces, so each piece has width ∆x =
L 10 = [f(20) + f(24) + f(28) + ... + f(52) + f(56)]∆x = [ln(20) + ln(24) + ln(28) + ... + ln(52) + ln(56)] · 4
∑^9
k=
ln(20 + 4k) · 4
