How do you find the inverse function of: $y=x^2+2x-15$ for $x>=-1$ ?

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Answer 1 · 2017-10-07T12:40:44

$color(blue)(f^-1(x)=sqrt(x-16)-1)$

$y = x^2+2x-15$

Get all terms containing variable on one side:

$x^2+2x = y+15$

Complete the square on the left side:

$(x^2+2x +(1)^2)-(1)^2 = y +15$

$(x^2+2x +(1)^2) = y +16$

$(x+1)^2= y+16$

Taking roots:

$x+1=+-sqrt(y-16)=> x = sqrt(y-16)-1 and x= -sqrt(y-16)-1$

$f^-1(x)=-sqrt(x-16)-1$ domain ${x in RR : 16<=x}$

$f^-1(x)=sqrt(x-16)-1$ domain ${x in RR : 16<=x}$

Range of $color(blue)(f^-1(x)=sqrt(x-16)-1)$

$color(blue)([-1 , oo))$

Range of $f^-1(x)=-sqrt(x-16)-1$

$(-oo , -1 ]$

How do you find the inverse function of: y=x^2+2x-15y=x^2+2x-15 for x>=-1x>=-1?