Finding function through a graph
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My understanding about checking if the function value lies on the graph is putting the x value and checking if it lies on the graph. Now in this example I'm a bit confused, when I solve a b and c, a gives function which will cover values greater than 1, and c will do the reverse. I think b option will cover the values in both + and -ve x axis. Am I interpreting it correctly?
linear-algebra functional-analysis functions arithmetic
edited Nov 30 '18 at 20:20
Misha Lavrov
44.4k555106
asked Nov 30 '18 at 18:14
shawn kshawn k
396
$endgroup$
|
show 3 more comments
$begingroup$
My understanding about checking if the function value lies on the graph is putting the x value and checking if it lies on the graph. Now in this example I'm a bit confused, when I solve a b and c, a gives function which will cover values greater than 1, and c will do the reverse. I think b option will cover the values in both + and -ve x axis. Am I interpreting it correctly?
linear-algebra functional-analysis functions arithmetic
edited Nov 30 '18 at 20:20
Misha Lavrov
44.4k555106
asked Nov 30 '18 at 18:14
shawn kshawn k
396
$endgroup$
2
$begingroup$
Note that $f(0) = 0$, which of the options satisfy this constraint?
$endgroup$
– caverac
Nov 30 '18 at 18:17
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@caverac none of these first 3 options are satisfying this condition!? then?
$endgroup$
– shawn k
Nov 30 '18 at 18:20
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$(B)$ clearly satisfies that.
$endgroup$
– KM101
Nov 30 '18 at 18:21
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Yes, one of them does. What I mean is, when you evaluate the function at $x = 0$, the result should be $0$
$endgroup$
– caverac
Nov 30 '18 at 18:21
1
$begingroup$
Option $(B)$ is what is known as a piecewise function (really just an absolute value function in this case). At $x = 0$, $x < 3$ so you use $y = -x$.
$endgroup$
– KM101
Nov 30 '18 at 18:22
|
show 3 more comments
$begingroup$
My understanding about checking if the function value lies on the graph is putting the x value and checking if it lies on the graph. Now in this example I'm a bit confused, when I solve a b and c, a gives function which will cover values greater than 1, and c will do the reverse. I think b option will cover the values in both + and -ve x axis. Am I interpreting it correctly?
linear-algebra functional-analysis functions arithmetic
edited Nov 30 '18 at 20:20
Misha Lavrov
44.4k555106
asked Nov 30 '18 at 18:14
shawn kshawn k
396
$endgroup$
My understanding about checking if the function value lies on the graph is putting the x value and checking if it lies on the graph. Now in this example I'm a bit confused, when I solve a b and c, a gives function which will cover values greater than 1, and c will do the reverse. I think b option will cover the values in both + and -ve x axis. Am I interpreting it correctly?
linear-algebra functional-analysis functions arithmetic
linear-algebra functional-analysis functions arithmetic
edited Nov 30 '18 at 20:20
Misha Lavrov
44.4k555106
asked Nov 30 '18 at 18:14
shawn kshawn k
396
edited Nov 30 '18 at 20:20
Misha Lavrov
44.4k555106
asked Nov 30 '18 at 18:14
shawn kshawn k
396
edited Nov 30 '18 at 20:20
Misha Lavrov
44.4k555106
edited Nov 30 '18 at 20:20
Misha Lavrov
44.4k555106
edited Nov 30 '18 at 20:20
Misha Lavrov
44.4k555106
44.4k555106
asked Nov 30 '18 at 18:14
shawn kshawn k
396
asked Nov 30 '18 at 18:14
shawn kshawn k
396
asked Nov 30 '18 at 18:14
shawn kshawn k
396
396
2
$begingroup$
Note that $f(0) = 0$, which of the options satisfy this constraint?
$endgroup$
– caverac
Nov 30 '18 at 18:17
$begingroup$
@caverac none of these first 3 options are satisfying this condition!? then?
$endgroup$
– shawn k
Nov 30 '18 at 18:20
$begingroup$
$(B)$ clearly satisfies that.
$endgroup$
– KM101
Nov 30 '18 at 18:21
$begingroup$
Yes, one of them does. What I mean is, when you evaluate the function at $x = 0$, the result should be $0$
$endgroup$
– caverac
Nov 30 '18 at 18:21
1
$begingroup$
Option $(B)$ is what is known as a piecewise function (really just an absolute value function in this case). At $x = 0$, $x < 3$ so you use $y = -x$.
$endgroup$
– KM101
Nov 30 '18 at 18:22
|
show 3 more comments
2
$begingroup$
Note that $f(0) = 0$, which of the options satisfy this constraint?
$endgroup$
– caverac
Nov 30 '18 at 18:17
$begingroup$
@caverac none of these first 3 options are satisfying this condition!? then?
$endgroup$
– shawn k
Nov 30 '18 at 18:20
$begingroup$
$(B)$ clearly satisfies that.
$endgroup$
– KM101
Nov 30 '18 at 18:21
$begingroup$
Yes, one of them does. What I mean is, when you evaluate the function at $x = 0$, the result should be $0$
$endgroup$
– caverac
Nov 30 '18 at 18:21
1
$begingroup$
Option $(B)$ is what is known as a piecewise function (really just an absolute value function in this case). At $x = 0$, $x < 3$ so you use $y = -x$.
$endgroup$
– KM101
Nov 30 '18 at 18:22
2
2
$begingroup$
Note that $f(0) = 0$, which of the options satisfy this constraint?
$endgroup$
– caverac
Nov 30 '18 at 18:17
$begingroup$
Note that $f(0) = 0$, which of the options satisfy this constraint?
$endgroup$
– caverac
Nov 30 '18 at 18:17
$begingroup$
@caverac none of these first 3 options are satisfying this condition!? then?
$endgroup$
– shawn k
Nov 30 '18 at 18:20
$begingroup$
@caverac none of these first 3 options are satisfying this condition!? then?
$endgroup$
– shawn k
Nov 30 '18 at 18:20
$begingroup$
$(B)$ clearly satisfies that.
$endgroup$
– KM101
Nov 30 '18 at 18:21
$begingroup$
$(B)$ clearly satisfies that.
$endgroup$
– KM101
Nov 30 '18 at 18:21
$begingroup$
Yes, one of them does. What I mean is, when you evaluate the function at $x = 0$, the result should be $0$
$endgroup$
– caverac
Nov 30 '18 at 18:21
$begingroup$
Yes, one of them does. What I mean is, when you evaluate the function at $x = 0$, the result should be $0$
$endgroup$
– caverac
Nov 30 '18 at 18:21
1
1
$begingroup$
Option $(B)$ is what is known as a piecewise function (really just an absolute value function in this case). At $x = 0$, $x < 3$ so you use $y = -x$.
$endgroup$
– KM101
Nov 30 '18 at 18:22
$begingroup$
Option $(B)$ is what is known as a piecewise function (really just an absolute value function in this case). At $x = 0$, $x < 3$ so you use $y = -x$.
$endgroup$
– KM101
Nov 30 '18 at 18:22
|
show 3 more comments
1 Answer
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$begingroup$
From inspection, it is clear the graph passes through $(0, 0)$, the origin. Neither $y = vert 3x-3vert$ nor $y = vert3x-1vert$ satisfy this.
Looking at option $(B)$, at $x = 0$, you have $x < 3$, so $y = -x$, meaning $y = 0$. Hence, $(B)$ satisfies this condition. Checking the other two points, you can see both $(3, -3)$ and $(6, 0)$ satisfy $y = x-6$ (because $x geq 3$). Hence, option $(B)$ is correct.
This function is known known as a “piecewise function” since the function contains two “sub-functions” which apply for a certain part of the domain. If $x geq 3$, you have a different function than if $x < 3$ (and they’re independent of each other). The absolute value function is a piecewise function because $vert xvert = x$ if $x geq 0$ and $vert xvert = -x$ if $x < 0$, and option $(B)$’s function is actually the absolute value function $y = vert x-3vert -3$ but it’s written as two separate functions.
$$xgeq 3 implies x-3 geq 0 implies y = x-3-3 implies y = x-6$$
$$x < 3 implies x-3 < 0 implies y = -(x-3)-3 implies y = -x+3-3 implies y = -x$$
answered Nov 30 '18 at 18:29
KM101KM101
5,9061423
$endgroup$
$begingroup$
+1 for awesomeness!
$endgroup$
– shawn k
Nov 30 '18 at 18:35
add a comment |
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$begingroup$
From inspection, it is clear the graph passes through $(0, 0)$, the origin. Neither $y = vert 3x-3vert$ nor $y = vert3x-1vert$ satisfy this.
Looking at option $(B)$, at $x = 0$, you have $x < 3$, so $y = -x$, meaning $y = 0$. Hence, $(B)$ satisfies this condition. Checking the other two points, you can see both $(3, -3)$ and $(6, 0)$ satisfy $y = x-6$ (because $x geq 3$). Hence, option $(B)$ is correct.
This function is known known as a “piecewise function” since the function contains two “sub-functions” which apply for a certain part of the domain. If $x geq 3$, you have a different function than if $x < 3$ (and they’re independent of each other). The absolute value function is a piecewise function because $vert xvert = x$ if $x geq 0$ and $vert xvert = -x$ if $x < 0$, and option $(B)$’s function is actually the absolute value function $y = vert x-3vert -3$ but it’s written as two separate functions.
$$xgeq 3 implies x-3 geq 0 implies y = x-3-3 implies y = x-6$$
$$x < 3 implies x-3 < 0 implies y = -(x-3)-3 implies y = -x+3-3 implies y = -x$$
answered Nov 30 '18 at 18:29
KM101KM101
5,9061423
$endgroup$
$begingroup$
+1 for awesomeness!
$endgroup$
– shawn k
Nov 30 '18 at 18:35
add a comment |
$begingroup$
From inspection, it is clear the graph passes through $(0, 0)$, the origin. Neither $y = vert 3x-3vert$ nor $y = vert3x-1vert$ satisfy this.
Looking at option $(B)$, at $x = 0$, you have $x < 3$, so $y = -x$, meaning $y = 0$. Hence, $(B)$ satisfies this condition. Checking the other two points, you can see both $(3, -3)$ and $(6, 0)$ satisfy $y = x-6$ (because $x geq 3$). Hence, option $(B)$ is correct.
This function is known known as a “piecewise function” since the function contains two “sub-functions” which apply for a certain part of the domain. If $x geq 3$, you have a different function than if $x < 3$ (and they’re independent of each other). The absolute value function is a piecewise function because $vert xvert = x$ if $x geq 0$ and $vert xvert = -x$ if $x < 0$, and option $(B)$’s function is actually the absolute value function $y = vert x-3vert -3$ but it’s written as two separate functions.
$$xgeq 3 implies x-3 geq 0 implies y = x-3-3 implies y = x-6$$
$$x < 3 implies x-3 < 0 implies y = -(x-3)-3 implies y = -x+3-3 implies y = -x$$
answered Nov 30 '18 at 18:29
KM101KM101
5,9061423
$endgroup$
$begingroup$
+1 for awesomeness!
$endgroup$
– shawn k
Nov 30 '18 at 18:35
add a comment |
$begingroup$
From inspection, it is clear the graph passes through $(0, 0)$, the origin. Neither $y = vert 3x-3vert$ nor $y = vert3x-1vert$ satisfy this.
Looking at option $(B)$, at $x = 0$, you have $x < 3$, so $y = -x$, meaning $y = 0$. Hence, $(B)$ satisfies this condition. Checking the other two points, you can see both $(3, -3)$ and $(6, 0)$ satisfy $y = x-6$ (because $x geq 3$). Hence, option $(B)$ is correct.
This function is known known as a “piecewise function” since the function contains two “sub-functions” which apply for a certain part of the domain. If $x geq 3$, you have a different function than if $x < 3$ (and they’re independent of each other). The absolute value function is a piecewise function because $vert xvert = x$ if $x geq 0$ and $vert xvert = -x$ if $x < 0$, and option $(B)$’s function is actually the absolute value function $y = vert x-3vert -3$ but it’s written as two separate functions.
$$xgeq 3 implies x-3 geq 0 implies y = x-3-3 implies y = x-6$$
$$x < 3 implies x-3 < 0 implies y = -(x-3)-3 implies y = -x+3-3 implies y = -x$$
answered Nov 30 '18 at 18:29
KM101KM101
5,9061423
$endgroup$
From inspection, it is clear the graph passes through $(0, 0)$, the origin. Neither $y = vert 3x-3vert$ nor $y = vert3x-1vert$ satisfy this.
Looking at option $(B)$, at $x = 0$, you have $x < 3$, so $y = -x$, meaning $y = 0$. Hence, $(B)$ satisfies this condition. Checking the other two points, you can see both $(3, -3)$ and $(6, 0)$ satisfy $y = x-6$ (because $x geq 3$). Hence, option $(B)$ is correct.
This function is known known as a “piecewise function” since the function contains two “sub-functions” which apply for a certain part of the domain. If $x geq 3$, you have a different function than if $x < 3$ (and they’re independent of each other). The absolute value function is a piecewise function because $vert xvert = x$ if $x geq 0$ and $vert xvert = -x$ if $x < 0$, and option $(B)$’s function is actually the absolute value function $y = vert x-3vert -3$ but it’s written as two separate functions.
$$xgeq 3 implies x-3 geq 0 implies y = x-3-3 implies y = x-6$$
$$x < 3 implies x-3 < 0 implies y = -(x-3)-3 implies y = -x+3-3 implies y = -x$$
answered Nov 30 '18 at 18:29
KM101KM101
5,9061423
edited Dec 6 '18 at 6:42
answered Nov 30 '18 at 18:29
KM101KM101
5,9061423
answered Nov 30 '18 at 18:29
KM101KM101
5,9061423
answered Nov 30 '18 at 18:29
KM101KM101
5,9061423
5,9061423
$begingroup$
+1 for awesomeness!
$endgroup$
– shawn k
Nov 30 '18 at 18:35
add a comment |
$begingroup$
+1 for awesomeness!
$endgroup$
– shawn k
Nov 30 '18 at 18:35
$begingroup$
+1 for awesomeness!
$endgroup$
– shawn k
Nov 30 '18 at 18:35
$begingroup$
+1 for awesomeness!
$endgroup$
– shawn k
Nov 30 '18 at 18:35
add a comment |
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2
$begingroup$
Note that $f(0) = 0$, which of the options satisfy this constraint?
$endgroup$
– caverac
Nov 30 '18 at 18:17
$begingroup$
@caverac none of these first 3 options are satisfying this condition!? then?
$endgroup$
– shawn k
Nov 30 '18 at 18:20
$begingroup$
$(B)$ clearly satisfies that.
$endgroup$
– KM101
Nov 30 '18 at 18:21
$begingroup$
Yes, one of them does. What I mean is, when you evaluate the function at $x = 0$, the result should be $0$
$endgroup$
– caverac
Nov 30 '18 at 18:21
1
$begingroup$
Option $(B)$ is what is known as a piecewise function (really just an absolute value function in this case). At $x = 0$, $x < 3$ so you use $y = -x$.
$endgroup$
– KM101
Nov 30 '18 at 18:22