What is the inverse of f(x)=x^2-16,x>0 ?

The inverse of f(x)=x^2-16,x>0 is sqrt(x+16),-sqrt(x+16)

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inverse of f(x)=x^2-16,x>0

inverse

f (x) = x^{2} - 16, x > 0

x + 16, - x + 16

Solution steps

y = x^{2} - 16

Replace

x

with

y

x = y^{2} - 16

Solve for

y, x = y^{2} - 16

x + 16, - x + 16

c \cdot e^{- c x}

h (t) = arcsin (2 t)

\frac{1}{\frac{x}{3} - 14}

- tan (\frac{4 π}{9})

- lo g_{4} (x - 1) + 2

What is the inverse of f(x)=x^2-16,x>0 ?
The inverse of f(x)=x^2-16,x>0 is sqrt(x+16),-sqrt(x+16)