Sample questions for test 2

There may be other types of questions, but being able to do these should mean you will do well on this test.

using MTH229

Zeros

• The bisection method is used to solve $f(x) = 0$ with $f(x) = x^2-2$. The interval $[1,2]$ is the intial interval. What interval is found after the first step? The second? The third? Use rational numbers.

• For $f(x) = \sin(x)$, is $[\pi/4, 3\pi/4]$ a bracketing interval?

• Use fzeros to find the real roots of the polynomial $f(x) = x^5 - x + 1$.

f(x) = x^5 - x + 1
fzeros(f)

1-element Array{Real,1}:
 -1.1673

There is just one, as this graph suggests

plot(f, -2, 2)
plot!(zero, -2, 2)

• For f above, the commands fzeros(f) and fzeros(f, -2, 2) will use different algorithms. Do they produce the same answer?

Yes, basically the same

No, not even close. Why not?

• For the function $f(x) = x^3 - 9x^2 +26x - 24$ over the interval $[0,6]$ there are a few roots. For each, find a bracketing interval, [a,b], and use fzero(f, [a,ealb]) to find the roots.

f(x) = x^3 - 9x^2 + 26x - 24
plot(f, 0,6)

Is [0,6] a bracketing interval?

Yes

No

What is the answer?

• Use fzero to find a solution to $\exp(x) = x^4$ in $[5, 15]$.

• Use fzero to find when the lines $y=100x + 200$ and $y= 50x + 300$ cross? (The $x$ value)

• Use fzeros to show numerically that $e^{x^2} + x \geq e^x$. Why can't you use fzero for this task?

Solving for when $e^{x^2} + x - e^x = 0$ with fzeros gives only one root. Since the function does not cross the $x$-axis, no bracketing interval can be found

Solving for when $e^{x^2} + x - e^x = 0$ with fzeros gives two roots. We could have found a bracket to find either one.

• Use fzeros to find how many zeros there are for $f(x) = \sin(20\pi\cdot x)$ over $[0,1]$.

(Is 1 counted in your answer?)

Limits

• Graphically identify the following limits or say why they do not exist:

$$~ \lim_{x \rightarrow 0} \frac{\sqrt{x + 25} - 5}{x} ~$$

The limit does not exist, there is a vertical asymptote

About 0.1

About 1.0

$$~ \lim_{x \rightarrow 0} \frac{|x|}{x} ~$$

About -1.0

About 1.0

The limit does not exist, the left and right limits are unequal.

$$~ \lim_{x \rightarrow 0} \frac{(3+x)^{-1} - (3-x)^{-1}}{x} ~$$

The limit is about 9

The limit is about -0.222

The limit does not exist, as `f(0) is an error.`

$$~ \lim_{x \rightarrow 0+} \log(\frac{1}{|x|})^{-1/32} ~$$

The limit looks like 0.8

The limit looks like 1

We can't really tell from the graph. As we plot closer around $0$, the value seems to get smaller, but doesn't converge to quickly

The limit looks like 0.9

• Using a table, identify the following limits. (You can use lim or just look at $f(0.1), f(0.01), f(0.001), ...$)

$$~ \lim_{x \rightarrow 0} \frac{\sqrt{x + 25} - 5}{x} ~$$

$$~ \lim_{x \rightarrow 0} \frac{|x|}{x} ~$$

The limit is -1

The limit does not exist

The limit is 1

$$~ \lim_{x \rightarrow 0} \frac{(3+x)^{-1} - (3-x)^{-1}}{x} ~$$

$$~ \lim_{x \rightarrow 0+} \log(\frac{1}{|x|})^{-1/32} ~$$

The limit is 0.649382

The values for lim have not converged by 1e-6. We can't really tell if the limit exists, and if it does what it is

• Using SymPy (the limit function) to find the following limits:

$$~ \lim_{x \rightarrow 0}\frac{\sqrt{1+x} - \sqrt{1-x}}{x}. ~$$

$$~ \lim_{x \rightarrow 4} \frac{x^{3/2} - 8}{x-4} ~$$

(Use: f(x) = (x^(3//2) - 8) / (x - 4).)

$$~ \lim_{x \rightarrow 0} \frac{|x|}{x}. ~$$

limit(f, 0, dir="-") is -1 and limit(f, 0, dir="+") is 1, so there is no limit

limit(f, 0) is 1, so the answer is 1

$$~ \lim_{x \rightarrow 0+} \log(\frac{1}{|x|})^{-1/32} ~$$