What is the inverse of f(x)=25-x^2,x>= 0 ?

The inverse of f(x)=25-x^2,x>= 0 is sqrt(-x+25),-sqrt(-x+25)

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inverse of f(x)=25-x^2,x>= 0

inverse

f (x) = 25 - x^{2}, x \geq 0

- x + 25, - - x + 25

Solution steps

y = 25 - x^{2}

Replace

x

with

y

x = 25 - y^{2}

Solve for

y, x = 25 - y^{2}

- x + 25, - - x + 25

m

f (x) = \frac{3 x + 1}{4 x + 7}

f (x) = 1 x + 4.1

(x - 1)^{2} - 5

2 x^{2} - 12 x + 3

What is the inverse of f(x)=25-x^2,x>= 0 ?
The inverse of f(x)=25-x^2,x>= 0 is sqrt(-x+25),-sqrt(-x+25)