What is the inverse of f(x)=49-x^2,x>= 0 ?

The inverse of f(x)=49-x^2,x>= 0 is sqrt(-x+49),-sqrt(-x+49)

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inverse of f(x)=49-x^2,x>= 0

inverse

f (x) = 49 - x^{2}, x \geq 0

- x + 49, - - x + 49

Solution steps

y = 49 - x^{2}

Replace

x

with

y

x = 49 - y^{2}

Solve for

y, x = 49 - y^{2}

- x + 49, - - x + 49

sin (4 x^{4})

f (x) = \frac{( x + 1 )}{x - 5}

\frac{- 4}{5}

f (x) = \frac{5}{4 x - 6} + 5

What is the inverse of f(x)=49-x^2,x>= 0 ?
The inverse of f(x)=49-x^2,x>= 0 is sqrt(-x+49),-sqrt(-x+49)