Question

could someone please show me how to find f^-1 of y=x^2+6x+7 algebraicly?

Answer 1

it does not have an inverse unless you restrict the domain

Answer 2

solve for x
use completing the square

Answer 3

right!
\[y=(x^2+6x)+7=(x^2+6x+9)+7-9=(x+3)^2-2 \]
vertex is (-3,-2)
so say we restrict the domain to (-3,inf)
so we have a 1 to 1 function now
\[0=x^2+6x+7-y\]
\[x=\frac{-6 \pm \sqrt{36-4(7-y)}}{2}=\frac{-6 \pm \sqrt{4}\sqrt{9-(7-y)}}{2}\]
\[=-3 \pm \sqrt{2+y}\]
so we have either
\[f^{-1}(x)=-3+\sqrt{2+y}\]
or
\[f^{-1}(x)=-3-\sqrt{2+y}\]
but not both
to determine which one it is we need to think our function has domain (-3,inf)
so then domain of f inverse should have range (-3,inf)
so its the first once
like shoose a number like 7 plug it in
for the first one we get 0 the second one gives us -6
only the first is in the range
so
\[f^{-1}(x)=-3 + \sqrt{2+y}\]

Answer 4

oooh! completing the square! that was why I couldn't solve the equation

Answer 5

oops
so then domain of f inverse should have range (-3,inf)<this sentence is weird
it should say the range of f inverse is (-3,inf)

Answer 6

shoose lol
should be choose

Answer 7

wait i think its other way around
domain of f: All real
range of f: (2,inf)

domain of f^-1 (2,inf)
range of f^-1; All real

Answer 8

*(-2,inf)

Answer 9

if f is defined on all of R then f^{-1} does not exist

Answer 10

what i was doing above cow was restricting the domain of f
such that the function would be 1 to 1 and therefore have an inverse

Answer 11

ok then restrict domain of f to (-3,inf)
then range of f^-1 is (-3,inf)

which is what myininaya was saying

Answer 12

lol

Answer 13

its like you knew what i was going to say

Answer 14

yeah i got mixed up a little bit