Mathematics 133 – Intensive Calculus for Science 1 Solutions – Final Exam

I. The function g(x) is approximately linear since all differences between successive entries are −.71, so all the slopes are equal to \frac{−.71}{.2} = −3.55. Using the first point, the equation is g(x) − 2.30 = (−3.55)(x − .2), so

g(x) = −3.55x + 3.01

The function f(x) is therefore approximately exponential. (This can also be seen since the ratios between successive entries are approximately the same.) To find an approximate equation f(x) = ba^{x}, we can use the first two table entries, which say

\begin{aligned}2.30 & = ba^{.2} \\ 2.12 & = ba^{.4} \end{aligned}

Hence dividing the second equation by the first, we get

.92 = \frac{2.30}{2.12} = \frac{ba^{.4}}{ba^{.2}} = a^{.2} = a^{\frac{1}{5}} .

Taking fifth powers to solve for a: a ≐ (.92)^{5} ≐ .66. Then from 2.3 = b(.66)^{.2}, we get b ≐ \frac{2.3}{(.66)^{.2}} ≐ 2.5. Therefore f(x) ≐ (2.5)(.66)^{x}.

II. A) The graph looks like a sinusoid with average value C = −2, amplitude A = 2, and

period 4 = \frac{2π}{B}, so B = \frac{π}{2} . The equation is

y = A sin(Bx) + C = 2 sin (\frac{πx}{2}) − 2

B) The part of the graph for x with −1 ≤ x ≤ 1 meets horizontal lines y = c with −4 ≤ c ≤ 0 exactly once. Therefore the restricted function has an inverse function. The domain of f^{−1} is the range of f – the interval [−4, 0].

III. A) The graph y = f′(x) looks approximately like:

(Note: three zeroes at x = 1, 2, 3 where y = f(x) has horizontal tangents, zero slope as x → ±∞ since y = f(x) is approaching a horizontal asymptote, and f′ is positive where f is increasing, negative where f is decreasing.)

B) The graph y = f′′(x) looks approximately like: IV. A)

f′(x) = lim_{h→0} \frac{f(x + h) − f(x)}{h}

= lim_{h→0} \frac{(x + h)^{2} + 2(x + h) + 3 − x^{2} − 2x − 3}{h}

= lim_{h→0} \frac{2xh + h^{2} + 2h}{h}

= lim_{h→0} 2x + h + 2

= 2x + 2

B) By the power rule,

g′(x) = 14x^{6} + \frac{3}{2}x^{\frac{−1}{2}} + \frac{2}{3}x^{− \frac{4}{3}}

C) By the exponential and logarithm rules, plus the chain rule,

h′(x) = ln(5)5^{x} − \frac{4x^{3}}{x^{4} + 3}

D) By the quotient rule,

i′(x) = \frac{(sin(x) − 1)(− sin(x)) − (cos(x))(cos(x))}{(sin(x) − 1)^{2}} = \frac{1}{sin(x) − 1}

E) By the rule for arctangent and the chain rule,

j′(x) = \frac{1}{1 + (3x + 2)^{2}} · 3 = \frac{3}{9x^{2} + 12x + 5}

A) The temperature at time t = 0 is

F (0) = 70 − 50e^{−0/10} = 70 − 50 = 20

degrees, B) The rate at which the temperature is changing at t = 10 is the derivative F′(10). By

the chain rule, F′(t) = −50e^{−t/10} · \frac{−1}{10} = 5e^{−t/10}. So F′(10) = 5e^{−1} ≐ 1.8 degrees per minute. (This is positive since the frozen dinner is warming up as it sits in the shopping cart.)

C) Solve 60 = 70 − 50e^{−t/10} for t:

\begin{aligned}−10 & = −50e^{−t/10} \\ .2 & = e^{−t/10} \\ ln(.2) & = −t/10 \\ −10 ln(.2) & = t \end{aligned}

Thus t = −10 ln(.2) ≐ 16 minutes.

VI. A) f is increasing on intervals where f′ is positive and decreasing on intervals where f′

is negative. Increasing: all x with a < x < c; decreasing: x < a and x > c. B) The critical points of f are the points where f′(x) = 0 – x = a, c. C) y = f(x) is concave up when f′(x) is increasing and concave down when f′(x) is

decreasing. Concave up: all x < b and x > d; concave down for all x with b < x < d.

VII. Suppose the pipeline is laid from the island to a point x miles along the shore toward the town, then along the shore to the town. The length under water is found by the Pythagorean theorem: \sqrt{x^{2} + 4}. The length along the shore is then 6 − x. So the cost is

C(x) = 3\sqrt{x^{2} + 4} + 2(6 − x).

To minimize this, we take the derivative (using the chain rule), set equal to zero, and solve for x:

0 = C′(x)

= \frac{3}{2}(x^{2} + 4)^{\frac{−1}{2}} (2x) − 2

= \frac{3x}{\sqrt{x^{2} + 4}} − 2

. Taking the 2 to the other side of the equation and clearing denominators, this gives:

3x = 2\sqrt{x^{2} + 4}

Square both sides: 9x^{2} = 4x^{2} + 16, so 5x^{2} = 16, and x = \sqrt{16/5} ≐ 1.7. The minimum cost pipeline joins the island to a point approximately 1.7 miles along the shoreline toward the town, then proceeds along the shoreline to the town.

VIII. A) The left and right-hand sums are

LHS = f(0) · \frac{1}{2} + f(\frac{1}{2}) · \frac{1}{2} + f(1) · \frac{1}{2} + f(\frac{3}{2}) · \frac{1}{2} + f(2) · \frac{1}{2} + f(\frac{5}{2}) · \frac{1}{2} + f(3) · \frac{1}{2}

+ f(\frac{7}{2}) · \frac{1}{2}

= 0 + \frac{1}{2} · \frac{1}{2} + 1 · \frac{1}{2} + \frac{3}{2} · \frac{1}{2} + 2 · \frac{1}{2} + 0 · \frac{1}{2} + (−2) · \frac{1}{2} + (−1) · \frac{1}{2}

= \frac{1}{4} + \frac{1}{2} + \frac{3}{4} + 1 − 1 − \frac{1}{2}

= 1

RHS = f(\frac{1}{2}) · \frac{1}{2} + f(1) · \frac{1}{2} + f(\frac{3}{2}) · \frac{1}{2}f(2) · \frac{1}{2} + f(\frac{5}{2}) · \frac{1}{2} + f(3) · \frac{1}{2} + f(\frac{7}{2}) · \frac{1}{2}

+ f(4) · \frac{1}{2}

= \frac{1}{2} · \frac{1}{2} + 1 · \frac{1}{2} + \frac{3}{2} · \frac{1}{2} + 2 · \frac{1}{2} + 0 · \frac{1}{2} + (−2) · \frac{1}{2} + (−1) · \frac{1}{2} + 0 · \frac{1}{2}

= 1

B) The integral is the area of the triangle formed by the graph over the interval [0, 5/2], minus the area of the triangle formed by the graph over the interval [5/2, 4] (since that lies below the x-axis). We have

∫_{0}^{4} f(x) dx = \frac{1}{2} · \frac{5}{2} · 2 − \frac{1}{2} · \frac{3}{2} · 2 = 1.