Question

Will give medal! please help!
what is the inverse of the given relation y=7x^2-3

Answer 1

\(\bf \color{red}{y}=7x^2-3\qquad inverse\implies x=7\color{red}{y}^2-3\)

notice, all we did was swap the variables about
then solve for "y"

Answer 2

ok yes i understand so far

Answer 3

so if you solve for "y", you'd get the inverse

Answer 4

would it be x=sqrt( y+3/7 ?

Answer 5

well.. you're meant to solve for "y"... the above shows a solution for "x"

Answer 6

oh sorry, like y^2= x+3/7?

Answer 7

.... wait.. a second

Answer 8

well... if we.... ahemm... follow that EXPRESSION, then we'd end up with \(\bf x=7y^2-3\implies \cfrac{x+3}{7}=y^2\implies \sqrt{\cfrac{x+3}{7}}=y\)

however, notice the graph of your original function --> http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiI3eF4yLTMiLCJjb2xvciI6IiNGMjE4MTgifSx7InR5cGUiOjEwMDB9XQ--
notice anything?

Answer 9

no im confused

Answer 10

well... is a parabola.. so.... the definition of an "inverse function" is obtained by the procedure we just did, yes, however, ONLY ONE-TO-ONE functions have an inverse function, and a parabola isn't a ONE-TO-ONE function, so by definition the original function, which is just a parabola, wouldn't have an inverse function

Answer 11

a one-to-one function has to pass the "vertical line test" as well as the "horizontal line test", a parabola passes the "vertical" one, not the "horizontal" one though, thus

Answer 12

umm , so your saying their is no inverse?

Answer 13

yes, because the original function is not a one-to-one function, it's a function nonetheless, just not a one-to-one, and by definition that's required for an inverse function to be

Answer 14

well their are answer options so their has to be one

Answer 15

well, the expression we'd get would be \(\bf x=7y^2-3\implies \cfrac{x+3}{7}=y^2\implies \sqrt{\cfrac{x+3}{7}}=y\)

Answer 16

here:

Answer 17

ohhh smokes... .I see the semantics.... ok... it says RELATION, no function.... ok then my bad

Answer 18

ohh ok

Answer 19

well, a root has a \(\bf \pm \) results, so you can always write that as \(\bf x=7y^2-3\implies \cfrac{x+3}{7}=y^2\implies \pm\sqrt{\cfrac{x+3}{7}}=y\)

Answer 20

ok yes, thank u!

Answer 21

yw