What is the inverse of f(x)=x^2-17,x>= 0 ?

The inverse of f(x)=x^2-17,x>= 0 is sqrt(x+17),-sqrt(x+17)

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inverse of f(x)=x^2-17,x>= 0

inverse

f (x) = x^{2} - 17, x \geq 0

x + 17, - x + 17

Solution steps

y = x^{2} - 17

Replace

x

with

y

x = y^{2} - 17

Solve for

y, x = y^{2} - 17

x + 17, - x + 17

f (x) = \frac{5 x ^{3}}{2}

f (x) = \frac{200 x}{400 - x}

\frac{9 - x ^{2}}{4}

f (x) = \frac{2 ^{x}}{5 + 2 ^{x}}

y = - (x + 3)^{2}

What is the inverse of f(x)=x^2-17,x>= 0 ?
The inverse of f(x)=x^2-17,x>= 0 is sqrt(x+17),-sqrt(x+17)