I have a question about distribution and absolute values. I was solving a problem and was wondering if it would be okay to distribute a number into an absolute value with two terms. For example $3|2x+3|+3x^2-5$, is it okay to distribute the $3$ into $|2x+3|$ to get $|6x+9|$? I have searched this on Google but people said it was not okay to do it, but someone else said it was okay to do it. I am really confused, could someone clear this up for me?
[Math] Distribution and Absolute Value
absolute value
Related Solutions
To solve these problems, you need to be aware of the geometry of the normal curve, and what the numbers in your standard normal table represent. Sometimes, they don't directly give you the number you want, and some manipulation is needed.
$1.$ I think the intention is to say let $Z$ be standard normal. Find $a$ so that $\Pr(|Z|\lt a)= .383$.
So the probability that $-a\lt Z\lt a$ is $.383$. Think of the standard normal curve. By symmetry, we want $-a\lt Z\lt 0=\Pr(0\lt Z\lt a=\frac{.383}{2}=.1915$.
So we want the area below $a$ to be $0.5+.1915=.6915$. If your table gives the probabilities that $Z\lt z$, look up $0.6915$ in the body of the table.
$2.$ The mean is $84$. You want $\Pr(|X-84|\gt 2.9$. This is the probability that $X\gt 84+2.9$ or less than $84-2.9$. So you want the probability that it differs from $84$ by more than $2.9$ in either direction, too low or too high. Find the individual probabilities and add. But by symmetry the two probabilities are the same. Can you find the probability that $X\gt 86.9$? If you do, double the result.
$3.$ The mean is $400$. We want the $k$ such that $\Pr(|X-400|\lt k)=0.975$. So In the two "tails", past $400+k$ and before $400-k$, we should have probability $0.025$. Thus each tail should have probability $0.0125$. What that means is that the total area below $400+k$ should be $1-0.0125=0.9875$. Perhaps you can take over from here.
$4.$ The probability that $35\lt X\lt 45$ is $\Pr(X\lt 45)-\Pr(X\lt 35)$. So the assertion that this probability is $0.65$ is just another way of saying that $\Pr(X\lt 45)=0.67$. Now to find $m$ and $s$, express the probability that $X\lt 35$ and the probability that $X\lt 45$ in terms of $m$ and $s$. It sounds as if you have done this for one of them. So it for the other and you will get $2$ linear equations in two unknowns. Solve.
In calculus courses one deals with the function $\ln\colon(0,+\infty)\to\mathbb R$. For $x>0$ it holds that $$ \frac{d}{dx}\ln x=\frac{1}{x}. $$ Note that the function $x\mapsto 1/x$ is defined for $x\in\mathbb R$ with $x\neq 0$. Integrating, one has $$ \int \frac{1}{x}\,dx= \begin{cases} \ln x+C_1 & x>0\\ \ln (-x)+C_2 & x<0. \end{cases} \tag{*} $$ This is sometimes written $$ \int\frac{1}{x}\,dx=\ln |x|+C $$ but in applications one consider either $x>0$ or $x<0$.
In complex analysis one also deals with a logarithm, taking complex numbers as arguments (I will not dig into the subject of branches here). One common definition is $$ \log z=\ln |z|+i\arg z. $$ Here $\ln$ is the logarithm defined in calculus. $\arg$ is the argument function. As it happens, also $$ \frac{d}{dz}\log z=\frac{1}{z}. $$ Here $\frac{d}{dz}$ is the complex derivative. Thus, in the complex setting $$ \int\frac{1}{z}\,dz=\log z+C, $$ i.e. the logarithm is still an antiderivative of $1/z$. Here, again, one should be more careful with domains, but since I understand it as OP is not into complex analysis yet (correct me if wrong), I do not dwell about that.
Finally, maple has no reason to expect the $x$ in $1/x$ to be real, so when you hit
int(1/x,x);
maple makes no assumption on $x$, but return
$\log x$.
Note that, if $x$ happens to be real and positive, and one assumes $\arg x=0$ (which is natural), then $$ \log x=\ln|x|+i\arg x=\ln x $$ so the different logarithms really coincide.
What should you do?
I suggest that you, every time you must integrate something that will return a logarithm, pause and think if you know anything about the variable. If you know that $x$ is positive, then $\int 1/x\,dx=\ln x+C$. If you know that $x$ is negative, then $\int 1/x\,dx=\ln(-x)+C=\ln|x|+C$ and if you do not know, then use $(*)$ above.
Best Answer
If the multiplier is non-negative, it's OK. For example, $2|x| = |2x|$ but $-3|x|=-|3x|$.