Inv functions (1 Viewer)

[Average Boreduser]

don't these result in -x given the restriction for f(x)?

I got the inv function to be -sqrt(x) basing it from the original graph of f(x)

I got an Idea on why it might be wrong (can ya'll confirm if I'm right?):
as f(x) is restricted to x<=0, we can simplify the function to sqrtx- when we find the inverse, f^-1(x) will = x^2. Hence, we can prove the two identities

^reason why I'm asking is bc I get abs(x) If I implement it, seems flawed

 

[Average Boreduser]

This is the qn

 

[Masaken]

don't these result in -x given the restriction for f(x)?

I got the inv function to be -sqrt(x) basing it from the original graph of f(x)

I got an Idea on why it might be wrong (can ya'll confirm if I'm right?):
as f(x) is restricted to x<=0, we can simplify the function to sqrtx- when we find the inverse, f^-1(x) will = x^2. Hence, we can prove the two identities

^reason why I'm asking is bc I get abs(x) If I implement it, seems flawed

the inverse function of x^2 with that restriction is -sqrt(x), i got that also and just to make sure i plugged it into desmos. i have no idea how you got an absolute value graph, what did you do?

 

[Average Boreduser]

I just realised the 'alternative' I used doesn't work either, theres surely no way to prove those identities

 

[Masaken]

I just realised the 'alternative' I used doesn't work either, theres surely no way to prove those identities

oh my bad, i forgot about the second part of the question. it is entirely possible to prove those identities:
for the first one, you sub x^2 into inverse f(x) with the restriction, so that when you simplify it you end up getting -(-x) = x
for the second one, when you square -sqrt x, you end up getting x.

 

[Average Boreduser]

wait hold on, when you find invf(f(x))= -sqrt(x^2)=-x isn't it -x?

 

[Masaken]

sorry for the crappy handwriting, i don't know how to use latex, but here's the working out for proving the first identity ^^
essentially, when you're finding the square root of x^2 with a negative in front of the entire root, you should get -x inside the root because x^2 has two square roots - x and -x, but you only take the negative solution because of the restriction. so you should be left with -x on the inside and another -ve sign on the front which cancel out to give x

 

[Average Boreduser]

Is the (-) in front bc f(x) was defined for x<=0?

 

[Masaken]

inverse f(x^2) means that you sub in x^2 (with the restriction) - which is f(x) - into the inverse function equation you found, which is -sqrt(x), so that means the x is replaced by x^2 but the -ve sign in the front of the square root remains there because it is not affected by the substitution (until you simplify the expression where the two negative signs cancel out)