What is the inverse of f(x)=7-x^2,x>= 0 ?

The inverse of f(x)=7-x^2,x>= 0 is sqrt(-x+7),-sqrt(-x+7)

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inverse of f(x)=7-x^2,x>= 0

inverse

f (x) = 7 - x^{2}, x \geq 0

- x + 7, - - x + 7

Solution steps

y = 7 - x^{2}

Replace

x

with

y

x = 7 - y^{2}

Solve for

y, x = 7 - y^{2}

- x + 7, - - x + 7

(10.1) (6.5)

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

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

r^{2} + 8.12 r + 3364

y = - x^{2} + 2 x - 4

What is the inverse of f(x)=7-x^2,x>= 0 ?
The inverse of f(x)=7-x^2,x>= 0 is sqrt(-x+7),-sqrt(-x+7)