Finding function through a graph












1












$begingroup$


enter image description here



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







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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
















1












$begingroup$


enter image description here



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







share|cite|improve this question











$endgroup$








  • 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














1












1








1





$begingroup$


enter image description here



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







share|cite|improve this question











$endgroup$




enter image description here



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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 20:20









Misha Lavrov

44.4k555106




44.4k555106










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














  • 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










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.



$$xgeq 3 implies x-3 geq 0 implies y = x-3-3 implies y = x-6$$



$$x < 3 implies x-3 < 0 implies y = -(x-3)-3 implies y = -x+3-3 implies y = -x$$






share|cite|improve this answer











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  • $begingroup$
    +1 for awesomeness!
    $endgroup$
    – shawn k
    Nov 30 '18 at 18:35











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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.



$$xgeq 3 implies x-3 geq 0 implies y = x-3-3 implies y = x-6$$



$$x < 3 implies x-3 < 0 implies y = -(x-3)-3 implies y = -x+3-3 implies y = -x$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    +1 for awesomeness!
    $endgroup$
    – shawn k
    Nov 30 '18 at 18:35
















2












$begingroup$

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.



$$xgeq 3 implies x-3 geq 0 implies y = x-3-3 implies y = x-6$$



$$x < 3 implies x-3 < 0 implies y = -(x-3)-3 implies y = -x+3-3 implies y = -x$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    +1 for awesomeness!
    $endgroup$
    – shawn k
    Nov 30 '18 at 18:35














2












2








2





$begingroup$

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.



$$xgeq 3 implies x-3 geq 0 implies y = x-3-3 implies y = x-6$$



$$x < 3 implies x-3 < 0 implies y = -(x-3)-3 implies y = -x+3-3 implies y = -x$$






share|cite|improve this answer











$endgroup$



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.



$$xgeq 3 implies x-3 geq 0 implies y = x-3-3 implies y = x-6$$



$$x < 3 implies x-3 < 0 implies y = -(x-3)-3 implies y = -x+3-3 implies y = -x$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 6:42

























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


















  • $begingroup$
    +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


















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