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

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Answer 1 · 2017-10-07T12:40:44

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

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

Get all terms containing variable on one side:

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

Complete the square on the left side:

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

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

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

Taking roots:

$x + 1 = \pm \sqrt{y - 16} \Rightarrow x = \sqrt{y - 16} - 1 and x = - \sqrt{y - 16} - 1$ $x+1=+-sqrt(y-16)=> x = sqrt(y-16)-1 and x= -sqrt(y-16)-1$

$f^{- 1} (x) = - \sqrt{x - 16} - 1$ $f^-1(x)=-sqrt(x-16)-1$ 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=x2+2x−15y=x^2+2x-15 for x>=-1x≥−1x>=-1?