Question

range:

Answer 1

\[y=\frac{ 8 }{ (4+x ^{2}) }\]
the book says tha range is \[0<y \le2\]
how come?

Answer 2

i tried to solve for x to find it but i got nothing..

Answer 3

cross multiply..
\[\implies y(4 + x^2) = 8\]
distribute
\[\implies 4y + x^2y = 8\]
isolate x
\[\implies x^2y = 8 - 4y\]
divide by y
\[\implies x^2 = \frac{8-4y}{y}\]
take the square root
\[\implies x = \sqrt{\frac{8-4y}{y}}\]
y cant be 0 because it will be undefined
y cant be greater than 2 because the radicand will be negative so \[0<y\le 2\]
does that help?

Answer 4

yes, thank you!
i see where i went wrong, i had changed the y to zero before multiplying :)

Answer 5

i see. glad you saw your mistake