What is the inverse of f(x)=4+x^2,x>= 0 ?

The inverse of f(x)=4+x^2,x>= 0 is sqrt(x-4),-sqrt(x-4)

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inverse of f(x)=4+x^2,x>= 0

inverse

f (x) = 4 + x^{2}, x \geq 0

x - 4, - x - 4

Solution steps

y = 4 + x^{2}

Replace

x

with

y

x = 4 + y^{2}

Solve for

y, x = 4 + y^{2}

x - 4, - x - 4

\frac{6 s + 3}{s ^{2}}

f (x) = \frac{1}{x ^{2} + 1}

f (x) = 12 x^{2} - 14 x

f (x) = (x - 3)^{0.5}

\frac{x - 2}{x + 2}

What is the inverse of f(x)=4+x^2,x>= 0 ?
The inverse of f(x)=4+x^2,x>= 0 is sqrt(x-4),-sqrt(x-4)