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i would like to clarify some questions from GRE,which at first seems a little difficult to understand,suppose that we have some function

$f(x)=|2*x|+4$ and graph of this function is given enter image description here

where is following question: For which of the following functions g defined for all numbers x does the graph of g intersect the graph of f ? and 4 possible answers are

A.g(x)=x-2
B.g(x)=x+3
C.g(x)=2*x-2
D.g(x)=2*x+3
E.g(x)=3*x-2

first what i did not understand what does mean

For which of the following functions g defined for all numbers x does the graph of g intersect the graph of f ? does it means that which g intersect of f for all x or?answer is E,i have guessed that for intersection we should have $|2*x|+4=g(x)$,clearly C and D no,because they are parallel lines with common slopes,so whe should have

$|2*x|+4=x-2$

or $|2*x|+4=x+3$

or $|2*x|+4=3*x-2$

after some calculation we will get $|2*x|=x-6$

or $|2*x|=x-3$

|2*x|=3*x-6

first can't be ,because $|2*x|>=x$ second also using the same rule,only one left

$|2*x|=3*x-6$ by using solving,for example if $x>0$ $2*x=3*x-6$ $x=6$

if $x<0$

$-2*x=3*x-6$

$x=6/5$

is there any short way to solve it?

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2  
I would just think of it geometrically. Look at the $y$-intercept and slope of one of the lines (the graph of $g$ will intersect the graph of $f$ if $g(0)<4$ and the slope of $y=g(x)$ is greater than $2$ or less than $-2$, for example). It seems you used similar reasoning for some parts; but you can do it for all of them. You don't want to waste time doing needless calculation... –  David Mitra Jun 20 '13 at 14:06
    
this criteria is because of modulus operator? –  dato datuashvili Jun 20 '13 at 14:09
1  
If I understand you correctly, yes. You have to consider "both sides" of the graph of $f$. For example, in E., the graph of $g$ will intersect the graph of $f$ at some $x>0$, since $g(0)<f(0)$ and the slope of $y=g(x)$ is greater than the slope of the "right side of $f$". (Sorry for the sloppy phrasing). If there were a part F.: $g(x)=-3x-2$, then the graphs wold intersect for some $x<0$. Perhaps sketching the graphs would help... –  David Mitra Jun 20 '13 at 14:14
    
ok thanks very much,i think that during the GRE test there will not be required too much calculation –  dato datuashvili Jun 20 '13 at 14:17
    
To put it simply, the graph of $f$ is a V shape, and each of the possible answers is a line. Choices A and B would look like $\underline{V}$, just with different amounts of space between the V and the underline, and choices C and D would look like V /, again just with different spacing between the V and the /. Only in choice E will the slash cut through the vee. –  cobaltduck Jun 20 '13 at 14:29

1 Answer 1

up vote 2 down vote accepted

You only need the two graphs to intersect at one point. If this is so for a single value of $x$, then the corresponding item will be a solution.

You are correct that "E" is the only answer. A quick way to solve the problem is to note that the graphs of all the items are straight lines; so you can deduce things from geometric reasoning. Just look at the $y$-intercepts of the lines and compare the slopes of the lines to the slopes of the line segments comprising the graph of $f$. It would also be helpful, for each item, to separate the problem into two, natural, parts: Does the given line intersect the graph of $f$ at some $x\ge0$? Does the given line intersect the graph of $f$ at some $x<0$?

For example:

E: The $y$ coordinate of the $y$-intercept of the line here is $-2$, which is less than $f(0)=4$, and the slope is $3$. Since the slope of the line segment given by $y=f(x), x\ge 0$ is $2<3$, the graph of $y=g(x)$ will intersect the graph of $y=f(x)$ for some $x>0$ (over $[0,\infty)$, the graph of $g$ rises more quickly than the graph of $f$).

If you sketch the graphs of the given lines (which you should become adept at for the GRE), paying close attention to the slope of the line, you can easily see what's happening.

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