What is the inverse of f(x)=sqrt(x^2-25),x>= 5 ?

The inverse of f(x)=sqrt(x^2-25),x>= 5 is sqrt(x^2+25),-sqrt(x^2+25)

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inverse of f(x)=sqrt(x^2-25),x>= 5

inverse

f (x) = x^{2} - 25, x \geq 5

x^{2} + 25, - x^{2} + 25

Solution steps

y = x^{2} - 25

Replace

x

with

y

x = y^{2} - 25

Solve for

y, x = y^{2} - 25

x^{2} + 25, - x^{2} + 25

\frac{x ^{2}}{x ^{2} - 9}

\frac{6 x}{5 x - 6}

y = cos (2 x)

\frac{x + 7}{x ^{2} - 14 x + 49}

csc (0.1 x + 1.2)

What is the inverse of f(x)=sqrt(x^2-25),x>= 5 ?
The inverse of f(x)=sqrt(x^2-25),x>= 5 is sqrt(x^2+25),-sqrt(x^2+25)