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i am asked to draw the graph of this min and max functions. these functions are given. $f_1(x) := 2, \quad f_2(x) := 3x, \quad f_3(x) := x^2$
now i need to draw the graph of these functions:

$g_1(x) := \min\{|f_1(x)|, |f_2(x)|\}, \quad g_2(x) := \max\{|f_1(x)|, |f_2(x)|\}$

the image and range of functions are $I \rightarrow \mathbb{R}$ with $I:=[-5,5]$

my question is, do i draw the whole functions graph even if i am asked for min or max? or do i draw just the min or max point?

thanks a lot

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3 Answers 3

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If we write $f(x)=\min (h(x),g(x))$, we are referring to the function whose value at any $x$ is the minimum of the two values of $h(x)$ and $g(x)$, so you should sketch the graphs of the functions.

For example, if $h(x)=0$ for $x<0$, $h(x)=1$ for $x\geq 0$, $g(x)=1$ for $x<0$ and $g(x)=0$ for $x\geq 0$ then $f(x)=0$ for all $x$.

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  • $\begingroup$ Thanks Jasper. so you mean the both graphs of h and g, right? $\endgroup$
    – doniyor
    Commented Dec 9, 2012 at 6:14
  • $\begingroup$ oh okay thanks much $\endgroup$
    – doniyor
    Commented Dec 9, 2012 at 6:31
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You’re to draw the graphs of the functions $g_1$ and $g_2$; both are functions on the domain $[-5,5]$, so each of your graphs should show the value of its function at each point of that domain.

The definition of $g_1$, for instance, means that at each point $x\in[-5,5]$, $$g_1(x)=\min\{|2|,|3x|\}=\min\{2,3|x|\}\;,$$ so you need to figure out what the graph of this function looks like. HINT: What’s different about the two cases $|x|\le\frac23$ and $\frac23\le|x|\le 5$?

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  • $\begingroup$ thanks Brian, this is what i was hesitating on... great $\endgroup$
    – doniyor
    Commented Dec 9, 2012 at 6:11
  • $\begingroup$ @doniyor: You’re welcome! $\endgroup$ Commented Dec 9, 2012 at 6:12
  • $\begingroup$ the first case is the min and second case is the max, right? but how did you come to $\frac{2}{3}$ $\endgroup$
    – doniyor
    Commented Dec 9, 2012 at 6:13
  • $\begingroup$ @doniyor: No, for each of the functions $g_1$ and $g_2$ you’ll need to look at both cases. For both functions you need to know at each $x\in[-5,5]$ which of $f_1(x)$ and $f_2(x)$ is bigger, so the obvious first step is to see where they’re equal, and $2=|3x|$ when $|x|=\frac23$. $\endgroup$ Commented Dec 9, 2012 at 6:16
  • $\begingroup$ oh yeah you are right $\endgroup$
    – doniyor
    Commented Dec 9, 2012 at 6:30
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For each $x$ in $[-5,5]$, $g_1 (x)$ will be whichever is lower: $|f_1(x)|$ or $|f_2(x)|$. What you might find useful is to plot both $|f_1|$ and $|f_2|$ on the same graph, and it will be clear at any point which is lower. This however, will not be the answer per se, because the graph of $y=g_1(x)$ is not the same as the two graphs together.

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