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 ]#