What is the inverse of f(x)=sqrt(-1+x),x>= 1 ?

The inverse of f(x)=sqrt(-1+x),x>= 1 is x^2+1

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inverse of f(x)=sqrt(-1+x),x>= 1

inverse

f (x) = - 1 + x, x \geq 1

x^{2} + 1

Solution steps

y = - 1 + x

Replace

x

with

y

x = - 1 + y

Solve for

y, x = - 1 + y

x^{2} + 1

f (x) = ln (2 x) - 3

g (x) = 2 x + 4

f (x) = - e^{x}

f (x) = \frac{( x ^{\frac{1}{7}} + 5 ) ^{3}}{7}

\frac{3 - 8 x}{4 + 7 x}

What is the inverse of f(x)=sqrt(-1+x),x>= 1 ?
The inverse of f(x)=sqrt(-1+x),x>= 1 is x^2+1