What is the inverse of 9-x^2,0<= x<= 3,x=2 ?

The inverse of 9-x^2,0<= x<= 3,x=2 is sqrt(-x+9),-sqrt(-x+9)

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inverse of 9-x^2,0<= x<= 3,x=2

inverse

9 - x^{2}, 0 \leq x \leq 3, x = 2

- x + 9, - - x + 9

Solution steps

y = 9 - x^{2}

Replace

x

with

y

x = 9 - y^{2}

Solve for

y, x = 9 - y^{2}

- x + 9, - - x + 9

f (x) = x + 4 = x^{2}

f (x) = 5 x - 50

k (x) = \frac{x - 1}{x + 7}

ln (0.25)

f (x) = 4 x - 5

What is the inverse of 9-x^2,0<= x<= 3,x=2 ?
The inverse of 9-x^2,0<= x<= 3,x=2 is sqrt(-x+9),-sqrt(-x+9)