Inverse Trig Problem (1 Viewer)

[Ostentatious]

Q22. of Excerise 26(a) of 3U Fitzpatrick.

I'm having trouble finding the inverse function of f(x)=x²+2x, x>2

My working:

Let f(x) = y <--- just for simplicity
Interchange x's and y's:
x=y²+2y
y²=x-2y
y=(x-2y)^1/2

STUCK. How can I eliminate the other y to return a function of y=-1+(1+x)^1/2 ?

Any help appreciated

 

[gurmies]

y = x^2 + 2x

= (x+1)^2 - 1

y + 1 = (x+1)^2

Interchanging x's and y's,

x + 1 = (y+1)^2

y + 1 = ±(x+1)^(1/2)

y = ±(x+1)^(1/2) - 1

 

[Ostentatious]

Thanks! I never would have thought of completing the square there.

 

[Pwnage101]

PS. this is an 'inverse' problem, NOT an 'inverse trig' problem

also could have used quadratic formula to find y in terms of x....

 

[gurmies]

No problem. This square was just dying to be completed :rofl:

 

[GUSSSSSSSSSSSSS]

dont forget the restriction: x > 2 so it only gonna be the positive square root xD

No problem. This square was just dying to be completed :rofl:

LOLLLLL =P

 

[Ostentatious]

No problem. This square was just dying to be completed :rofl:

I just couldn't deliver V_V

PS. this is an 'inverse' problem, NOT an 'inverse trig' problem

Sorry about that, right after that particular exercise the inverse trig begins. Plus, the chapter was called inverse trig so my head wasn't on right.

 

[Pwnage101]

Sorry about that, right after that particular exercise the inverse trig begins. Plus, the chapter was called inverse trig so my head wasn't on right.

no worries

 

[addikaye03]

PS. this is an 'inverse' problem, NOT an 'inverse trig' problem

also could have used quadratic formula to find y in terms of x....

That's a bit pety, don't you think?