JavaScript is disabled. Because that a better experience, please permit JavaScript in her browser prior to proceeding.

You are watching: Express y as a function of x


*

Reactions:1 person
*

$x - 8y^3 =0$add $8y^3$ come both political parties of the equation ...$x = 8y^3$divide both political parties by $8$ ...$dfracx8 = y^3$take the cube root of both sides ...$sqrt<3>dfracx8 = sqrt<3>y^3$$dfracsqrt<3>x2 = y$
Reactions:1 person
therefore you know that if (displaystyle x= y^2) you "invert" by making use of the square root: (displaystyle y= sqrtx= x^1/2). Well, if (displaystyle x= y^3) friend "invert" by utilizing the cube root: (displaystyle y= sqrt<3>x= x^1/3).
So you know that if (displaystyle x= y^2) you "invert" by making use of the square root: (displaystyle y= sqrtx= x^1/2).
But (displaystyle x = y^2 ) isn"t a function, and also it can"t it is in rewritten together a duty of x, which is the object of the OP"s thread.(displaystyle x = y^2, y ge 0 ) is a function, and it have the right to be rewritten together (displaystyle y= sqrtx= x^1/2).
But (displaystyle x = y^2 ) isn"t a function, and also it can"t be rewritten together a role of x, which is the topic of the OP"s thread.(displaystyle x = y^2, y ge 0 ) is a function, and it have the right to be rewritten together (displaystyle y= sqrtx= x^1/2).
$x=y^2$ is a function of $y$. Perhaps you mean that the is no a function of $x$. Or probably the term girlfriend are trying to find is the it is not a bijective function. Bijectivity is the actual requirement for a duty to have an inverse.
So you recognize that if (displaystyle x= y^2) you "invert" by utilizing the square root
: (displaystyle y= sqrtx= x^1/2).Well, if (displaystyle x= y^3) friend "invert" by making use of the cube root: (displaystyle y= sqrt<3>x= x^1/3).
But (displaystyle x = y^2 ) isn"t a function, and also it can"t it is in rewritten as a duty of x, i beg your pardon is the object of the OP"s thread.(displaystyle x = y^2, y ge 0 ) is a function, and also it have the right to be rewritten together (displaystyle y= sqrtx= x^1/2).
$x=y^2$ is a role of $y$. Possibly you average that that is not a function of $x$. Or maybe the term friend are in search of is that it is no a bijective function. Bijectivity is the actual necessity for a function to have an inverse.

See more: Diamond Supply Co X Travis Scott Diamond Shirt, Diamond Supply Co X Travis Scott Explosion T


If us are having a disagreement around functions need to we not have a clear concept what a function is? that is a question foundations class.Definition: intend that every of $A~&~B$ is a nonempty set.The statement that $f$ is a role from $A$ to $B$ method that:$ eginalign*&ullet~fsubseteq A imes B.\&ullet~ extevery xin A ext is the an initial term in precisely one pair in f\ endalign*$Now if $A=B=mathbbR$ it must be clear the $f=(x,x^2)$ is a role and $g=(x^2,x)$ is not!Also $f=(x,x^3)~&~g=(x^3,x)$ room both functions.
Similar mathematics DiscussionsMath ForumDate
how to express these integrals with attributes of Gamma or Beta?CalculusNov 22, 2018
How might one write an expression/'function' for these graphs?Pre-CalculusApr 22, 2018
How might one create an expression for a role that looks favor this?Pre-CalculusApr 16, 2018
expressing attributes as other functions to solve laplace transformsDifferential EquationsOct 18, 2017

Similar threads
How come express these integrals with attributes of Gamma or Beta?
How can one compose an expression/'function' because that these graphs?
How could one compose an expression because that a duty that looks prefer this?
expressing features as other attributes to settle laplace transforms

Math aid Forum

mathematics is involved with numbers, data, quantity, structure, space, models, and change. Established in 2005, Math help Forum is committed to cost-free math aid and math discussions, and our math ar welcomes students, teachers, educators, professors, mathematicians, engineers, and scientists.