Question

Define a function f : R → R by the formula
f(x) = 8x − 3.


Prove that f is onto.
Let y is in R. Then
y + 3
8
is in R
, and
f

y + 3
8

= 8



− 3
= .

Answer 1

f(x) = 8x − 3.


let y be in R
then y = 8((y+3)/8) -3 = f((y+3)/8)
so y = f(x) when x = (y+3)/8
qed

Answer 2

\[Let y \in R. Then \frac{ y+3 }{ 8 } \in R, and f(\frac{ y+3 }{ 8 }) = 8(?)-3 =\]

Answer 3

? = (y+3)/8

Answer 4

\[f(\frac{y+3}{8})= 8(\frac{y+3}{8})-3= y\]

Answer 5

\[-\infty\le f(x)\le \infty\]
\[-\infty\le (8x-3)\le \infty\]
\[-\infty\le 8x\le \infty\]
\[-\infty\le x\le \infty\]

Answer 6

prolly not a "correct" proof tho:)

Answer 7

that's cute:)

Answer 8

>zzr0ck3r, ok, that makes sense now that I see it

Answer 9

try and understand what @amistre64 did also because there is good insight there.
he showed that if f(x) that is real, we have an x that is real.

Answer 10

but any time Im showing onto
I let y be in R
write the inverse of the function f(x) = y, then composite that with the function

Answer 11

compose*

Answer 12

got more? These are fun and I need the brush up

Answer 13

quite a bit actually

Answer 14

sweet, ima grab coffee ill brb