What is the inverse of f(x) = [tex]\sqrt{x} \frac{1}{2}-5 + 2[/tex] for x [tex]\geq[/tex] 10
f−1(x)=12[(x+5)2−2]; x≥−5
f to the power of negative 1 end exponent left parenthesis x right parenthesis equals 1 half open square brackets open parentheses x plus 5 close parentheses squared minus 2 close square brackets, ; , x is greater than or equal to negative 5
f−1(x)=12[(x−2)2+5];x≤2
f to the power of negative 1 end exponent left parenthesis x right parenthesis equals 1 half open square brackets open parentheses x minus 2 close parentheses squared plus 5 close square brackets, ;, x is less than or equal to 2
f−1(x)=(2x+5)2−2; x≥−5
f to the power of negative 1 end exponent left parenthesis x right parenthesis equals open parentheses 2 x plus 5 close parentheses squared minus 2, ; , x is greater than or equal to negative 5
f−1(x)=2[(x−2)2+5];x≥
