trying to graph a function with x and e(constant ?)
$$ f(x)=begin{cases} 1-|x|/e ,quad |x| leqslant e \ 0, qquad qquad e<
|x| leqslant 1 end{cases} $$
$$ whereqquad 0< e < 1 $$
I assume e is a constant and e will not run from 0 to 1. But x will go from 0 to 1.
I attempted a graph trying to understand the function.
DATA:(x)
octave:23>
x =
Columns 1 through 7:
0.00000 0.05000 0.10000 0.15000 0.20000 0.25000 0.30000
Columns 8 through 14:
0.35000 0.40000 0.45000 0.50000 0.55000 0.60000 0.65000
Columns 15 through 19:
0.70000 0.75000 0.80000 0.85000 0.90000
=================================================
( y = 1 - x/.9) I took e= .9
so e is $$ 0<e<1 $$
DATA:(y)
octave:24>
y =
Columns 1 through 8:
1.00000 0.94444 0.88889 0.83333 0.77778 0.72222 0.66667 0.61111
Columns 9 through 16:
0.55556 0.50000 0.44444 0.38889 0.33333 0.27778 0.22222 0.16667
Columns 17 through 19:
0.11111 0.05556 0.00000
The graph of the function
algebra-precalculus
asked Nov 19 '18 at 2:06
tt z
51
add a comment |
$$ f(x)=begin{cases} 1-|x|/e ,quad |x| leqslant e \ 0, qquad qquad e<
|x| leqslant 1 end{cases} $$
$$ whereqquad 0< e < 1 $$
I assume e is a constant and e will not run from 0 to 1. But x will go from 0 to 1.
I attempted a graph trying to understand the function.
DATA:(x)
octave:23>
x =
Columns 1 through 7:
0.00000 0.05000 0.10000 0.15000 0.20000 0.25000 0.30000
Columns 8 through 14:
0.35000 0.40000 0.45000 0.50000 0.55000 0.60000 0.65000
Columns 15 through 19:
0.70000 0.75000 0.80000 0.85000 0.90000
=================================================
( y = 1 - x/.9) I took e= .9
so e is $$ 0<e<1 $$
DATA:(y)
octave:24>
y =
Columns 1 through 8:
1.00000 0.94444 0.88889 0.83333 0.77778 0.72222 0.66667 0.61111
Columns 9 through 16:
0.55556 0.50000 0.44444 0.38889 0.33333 0.27778 0.22222 0.16667
Columns 17 through 19:
0.11111 0.05556 0.00000
The graph of the function
algebra-precalculus
asked Nov 19 '18 at 2:06
tt z
51
add a comment |
$$ f(x)=begin{cases} 1-|x|/e ,quad |x| leqslant e \ 0, qquad qquad e<
|x| leqslant 1 end{cases} $$
$$ whereqquad 0< e < 1 $$
I assume e is a constant and e will not run from 0 to 1. But x will go from 0 to 1.
I attempted a graph trying to understand the function.
DATA:(x)
octave:23>
x =
Columns 1 through 7:
0.00000 0.05000 0.10000 0.15000 0.20000 0.25000 0.30000
Columns 8 through 14:
0.35000 0.40000 0.45000 0.50000 0.55000 0.60000 0.65000
Columns 15 through 19:
0.70000 0.75000 0.80000 0.85000 0.90000
=================================================
( y = 1 - x/.9) I took e= .9
so e is $$ 0<e<1 $$
DATA:(y)
octave:24>
y =
Columns 1 through 8:
1.00000 0.94444 0.88889 0.83333 0.77778 0.72222 0.66667 0.61111
Columns 9 through 16:
0.55556 0.50000 0.44444 0.38889 0.33333 0.27778 0.22222 0.16667
Columns 17 through 19:
0.11111 0.05556 0.00000
The graph of the function
algebra-precalculus
asked Nov 19 '18 at 2:06
tt z
51
$$ f(x)=begin{cases} 1-|x|/e ,quad |x| leqslant e \ 0, qquad qquad e<
|x| leqslant 1 end{cases} $$
$$ whereqquad 0< e < 1 $$
I assume e is a constant and e will not run from 0 to 1. But x will go from 0 to 1.
I attempted a graph trying to understand the function.
DATA:(x)
octave:23>
x =
Columns 1 through 7:
0.00000 0.05000 0.10000 0.15000 0.20000 0.25000 0.30000
Columns 8 through 14:
0.35000 0.40000 0.45000 0.50000 0.55000 0.60000 0.65000
Columns 15 through 19:
0.70000 0.75000 0.80000 0.85000 0.90000
=================================================
( y = 1 - x/.9) I took e= .9
so e is $$ 0<e<1 $$
DATA:(y)
octave:24>
y =
Columns 1 through 8:
1.00000 0.94444 0.88889 0.83333 0.77778 0.72222 0.66667 0.61111
Columns 9 through 16:
0.55556 0.50000 0.44444 0.38889 0.33333 0.27778 0.22222 0.16667
Columns 17 through 19:
0.11111 0.05556 0.00000
The graph of the function
algebra-precalculus
algebra-precalculus
asked Nov 19 '18 at 2:06
tt z
51
asked Nov 19 '18 at 2:06
tt z
51
asked Nov 19 '18 at 2:06
tt z
51
asked Nov 19 '18 at 2:06
tt z
51
asked Nov 19 '18 at 2:06
tt z
51
51
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
On $[0,e)$, it is a line segment $$y=1-frac{x}{e},$$
It has slope $-frac1e$ and intercept $1$.
Also, this is an even function, On $(-e,0]$, it is a line segment $$y=1+frac{x}{e},$$
It has slope $frac1e$ and intercept $1$.
It is zero everywhere else.
Remark: $e$ need not be a good choice of notation in general.
answered Nov 19 '18 at 2:11
Siong Thye Goh
99.3k1464117
so "|x|" suggest that 'x' goes in both direction , positive and negative ?
– tt z
Nov 19 '18 at 2:24
$|x|$ is the absolute value function. $|-0.5|=0.5=|0.5|$. It remove the sign and returns the length.
– Siong Thye Goh
Nov 19 '18 at 2:26
Thank you!!!! very much.
– tt z
Nov 19 '18 at 2:35
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
On $[0,e)$, it is a line segment $$y=1-frac{x}{e},$$
It has slope $-frac1e$ and intercept $1$.
Also, this is an even function, On $(-e,0]$, it is a line segment $$y=1+frac{x}{e},$$
It has slope $frac1e$ and intercept $1$.
It is zero everywhere else.
Remark: $e$ need not be a good choice of notation in general.
answered Nov 19 '18 at 2:11
Siong Thye Goh
99.3k1464117
so "|x|" suggest that 'x' goes in both direction , positive and negative ?
– tt z
Nov 19 '18 at 2:24
$|x|$ is the absolute value function. $|-0.5|=0.5=|0.5|$. It remove the sign and returns the length.
– Siong Thye Goh
Nov 19 '18 at 2:26
Thank you!!!! very much.
– tt z
Nov 19 '18 at 2:35
add a comment |
On $[0,e)$, it is a line segment $$y=1-frac{x}{e},$$
It has slope $-frac1e$ and intercept $1$.
Also, this is an even function, On $(-e,0]$, it is a line segment $$y=1+frac{x}{e},$$
It has slope $frac1e$ and intercept $1$.
It is zero everywhere else.
Remark: $e$ need not be a good choice of notation in general.
answered Nov 19 '18 at 2:11
Siong Thye Goh
99.3k1464117
so "|x|" suggest that 'x' goes in both direction , positive and negative ?
– tt z
Nov 19 '18 at 2:24
$|x|$ is the absolute value function. $|-0.5|=0.5=|0.5|$. It remove the sign and returns the length.
– Siong Thye Goh
Nov 19 '18 at 2:26
Thank you!!!! very much.
– tt z
Nov 19 '18 at 2:35
add a comment |
On $[0,e)$, it is a line segment $$y=1-frac{x}{e},$$
It has slope $-frac1e$ and intercept $1$.
Also, this is an even function, On $(-e,0]$, it is a line segment $$y=1+frac{x}{e},$$
It has slope $frac1e$ and intercept $1$.
It is zero everywhere else.
Remark: $e$ need not be a good choice of notation in general.
answered Nov 19 '18 at 2:11
Siong Thye Goh
99.3k1464117
On $[0,e)$, it is a line segment $$y=1-frac{x}{e},$$
It has slope $-frac1e$ and intercept $1$.
Also, this is an even function, On $(-e,0]$, it is a line segment $$y=1+frac{x}{e},$$
It has slope $frac1e$ and intercept $1$.
It is zero everywhere else.
Remark: $e$ need not be a good choice of notation in general.
answered Nov 19 '18 at 2:11
Siong Thye Goh
99.3k1464117
edited Nov 19 '18 at 2:27
answered Nov 19 '18 at 2:11
Siong Thye Goh
99.3k1464117
answered Nov 19 '18 at 2:11
Siong Thye Goh
99.3k1464117
answered Nov 19 '18 at 2:11
Siong Thye Goh
99.3k1464117
99.3k1464117
so "|x|" suggest that 'x' goes in both direction , positive and negative ?
– tt z
Nov 19 '18 at 2:24
$|x|$ is the absolute value function. $|-0.5|=0.5=|0.5|$. It remove the sign and returns the length.
– Siong Thye Goh
Nov 19 '18 at 2:26
Thank you!!!! very much.
– tt z
Nov 19 '18 at 2:35
add a comment |
so "|x|" suggest that 'x' goes in both direction , positive and negative ?
– tt z
Nov 19 '18 at 2:24
$|x|$ is the absolute value function. $|-0.5|=0.5=|0.5|$. It remove the sign and returns the length.
– Siong Thye Goh
Nov 19 '18 at 2:26
Thank you!!!! very much.
– tt z
Nov 19 '18 at 2:35
so "|x|" suggest that 'x' goes in both direction , positive and negative ?
– tt z
Nov 19 '18 at 2:24
so "|x|" suggest that 'x' goes in both direction , positive and negative ?
– tt z
Nov 19 '18 at 2:24
$|x|$ is the absolute value function. $|-0.5|=0.5=|0.5|$. It remove the sign and returns the length.
– Siong Thye Goh
Nov 19 '18 at 2:26
$|x|$ is the absolute value function. $|-0.5|=0.5=|0.5|$. It remove the sign and returns the length.
– Siong Thye Goh
Nov 19 '18 at 2:26
Thank you!!!! very much.
– tt z
Nov 19 '18 at 2:35
Thank you!!!! very much.
– tt z
Nov 19 '18 at 2:35
add a comment |
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