Some sample questions for test 1.

We need

using MTH229

calculator

Evaluate each expression. For each, write the answer and the julia commands you used to produce the answer:

• $10^9 / 8 \cdot 7 - 6 + 5$

10^9 / 8*7 - 6 + 5

• $\sin(\pi/3) \cdot \sin^2(\pi/4) \cdot \sin^{-1}(\pi/5)$

sin(pi/3) * sin(pi/4)^2 * asin(pi/5)

• $\cos(75^\circ) - \cos(75)$

cosd(75) - cos(75), cos(75*pi/180) - cos(75)

(-0.6629322246222286,-0.6629322246222286)

• $e^{1/2 \cdot(3 - 1)^2}$

exp(1/2 * (3-1)^2), e^(1/2*(3-1)^2)

(7.38905609893065,7.38905609893065)

*

$$~ \frac{\sin(x) + \cos(x)}{\sin(x) - \cos(x)}, \text{ when } x = 1. ~$$

x = 1
(sin(x) - cos(x)) / (sin(x) - cos(x))

*

$$~ \sin^2 \left(\frac{x - 2\pi}{x + 2\pi}\right) \text{ when } x = \pi/4. ~$$

x = pi/4
inner = (x-2pi) / (x + 2pi)
sin(inner)^2

Order of operations

Which of these three expressions will be different from the other two?

• 2/3/4, (2/3)/4, 2/(3/4)

2/3/4, (2/3)/4, 2/(3/4)  # left-right associativity, so last one

(0.16666666666666666,0.16666666666666666,2.6666666666666665)

• 4^3^2, (4^3)/2, 4^(3^2)

4^3^2, (4^3)/2, 4^(3^2)   # powers have right-left associativity, so middle

(262144,32.0,262144)

• 2 - 3 - 4, (2-3)-4, 2 - (3-4)

2 - 3 - 4, (2-3)-4, 2 - (3-4)  # most things-like subtraction-associate left right, so right one

(-5,-5,3)

Integers can cause some issues.

Are these expressions identical

• 2^(-1), 2.0^(-1)

ANS: 2^(-1) is an error. Integer bases and negative exponents are not defined (as the answer is expected to be an integer, not a rational, for technical reasons.)

• 2^64, 0

ANS: Yes (at least on 64 bit machines). This is due to integer wraparound, rather than checking for overflow. (Technical, and not really essential to know.)

Floating point can cause small issues.

Are these expressions identical?

• .1 + .2 - .3, 0

ANS: No. Try it. Why? You can't express .1, .2 and .3 exactly in floating point

• sin(pi), 0

ANS: No, not quite, but super close.

• cos(pi), -1.0

ANS: Yes

Defining functions

For each function, write a function in Julia and compute $f(1/2)$. For each, your answer should have the command to define the function and the value at $1/2$.

• $f(x) = x^4 - 4x^2 + 2$

f(x) = x^4 - 4x^2 + 2
f(1/2)

• $f(x) = \sqrt{x / (x + 1)}$

f(x) = sqrt(x / (x + 1))
f(1/2)

• $f(x) = \ln(x + 1) - x$

f(x) = log(x + 1) - x  ## NOT ln(x +1)!!!
f(1/2)

• $f(x) = \cos^{-1}(x)$

f(x) = acos(x)  # not cos^(-1)(x) which won't work at all!
f(1/2)

*

$$~ f(x) = \begin{cases} x^2 & x < 1\\ x & \text{otherwise} \end{cases} ~$$

f(x) = x < 1 ? x^2 : x
f(1/2)

• A function describing the line that goes through the points $(-1,-\pi)$ and $(2,\pi)$.

x0,y0 = -1, -pi
x1, y1 = 2, pi
m = (y1 - y0) / (x1 - x0)
f(x) =  y0 + m * (x - x0)  # point-slope form!
f(1/2)

Plotting

• Graph the function f(x) = x * (x-1) / (x -2) / (x-3). Are there vertical asymptotes? Is there a slant asymptote? Is there a horizontal asymptote?

f(x) = x * (x-1) / (x-2) / (x-3)
plot(f, -10, 10)

Can see VAs at 2, 3; HA seems clear and is 1

• Graph the function f(x) = cos(x) * cosh(x) between $[1, 5]$. Identify from the graph how many values of $x$ satisfy $f(x) = 1$ in $[1,5]$? What is the largest of them?

f(x) = cos(x) * cosh(x)
plot(f, 1, 5)
g(x) = 1
plot!(g, 1, 5)

Looks lke 1 out near 4.7.

• Graph the functions $f(x) = \sin(x)$ and $g(x) = x - x^3/6$ over the interval $[-\pi/2, \pi/2]$. Can you tell how many times the two functions cross? Does it help to plot $\sin(x) - (x - x^3/6)$? Would it be helpful to know that $\sin(x) - (x - x^3/6)$ is increasing on this interval?

f(x) = sin(x)
g(x) = x - x^3/6
plot(f, -pi/2, pi/2)
plot!(g, -pi/2, pi/2)

Does this help? Well let's plot the difference:

h(x) = f(x) - g(x)
plot(h, -pi/2, pi/2)

Since h is increasing it can only cross $0$ once, though it is hard to tell from a graph.

• For the function $f(x) = x^5 - x + 1$ identify any intervals where the function is decreasing within $[-5, 5]$.

f(x) = x^5 - x + 1
plot(f, -5, 5)

Well, clearly if $f$ is decreasing, it must happen between $-2.5$ and $2.5$:

plot(f, -2.5, 2.5)

Err, between -1 and 1!

plot(f, -1, 1)

That's better. It is between $[-0.67, 0.67]$.

• Two phone plans are given as follows: Plan 1 is 40 dollars per month for the first 5Gb of data and 2 dollars additional for any additional data usage; Plan 2 is a flat rate of 75 dollars per month, unlimited data. Graphically find the "break-even" point.

p1(x) = x < 5 ? 40 : 40 + 2(x-5)
p2(x) = 75
plot(p1, 0, 20)
plot!(p2, 0, 20)

Darn, not big enough. Let's adjust. It looks like for sure they will cross by $x=30$:

plot(p1, 0, 30)
plot!(p2, 0, 30)

And voila! Around 22.7