What is the inverse of f(x)=15-x^2,x>= 0 ?

The inverse of f(x)=15-x^2,x>= 0 is sqrt(-x+15),-sqrt(-x+15)

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inverse of f(x)=15-x^2,x>= 0

inverse

f (x) = 15 - x^{2}, x \geq 0

- x + 15, - - x + 15

Solution steps

y = 15 - x^{2}

Replace

x

with

y

x = 15 - y^{2}

Solve for

y, x = 15 - y^{2}

- x + 15, - - x + 15

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

f (x) = 8 x - 2

y = - 3 x + 2

1 + \frac{1}{x - 1}

\frac{2 x ^{2} - 8 x + 5}{( x - 2 ) ^{2}}

What is the inverse of f(x)=15-x^2,x>= 0 ?
The inverse of f(x)=15-x^2,x>= 0 is sqrt(-x+15),-sqrt(-x+15)