What is the inverse of f(x)=sqrt(x+1),0<= x<= 8 ?

The inverse of f(x)=sqrt(x+1),0<= x<= 8 is x^2-1

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inverse of f(x)=sqrt(x+1),0<= x<= 8

inverse

f (x) = x + 1, 0 \leq x \leq 8

x^{2} - 1

Solution steps

y = x + 1

Replace

x

with

y

x = y + 1

Solve for

y, x = y + 1

x^{2} - 1

arcsin (\frac{5}{3})

f (x) = - \frac{1}{4} x + \frac{3}{4}

f (x) = \frac{5 x}{4 x - 1}

\frac{5 + 13 x}{2}

f (x) = \frac{8 - x}{4}

What is the inverse of f(x)=sqrt(x+1),0<= x<= 8 ?
The inverse of f(x)=sqrt(x+1),0<= x<= 8 is x^2-1