What is the inverse of f(x)=2-x^2,sqrt(3)<= x<= 2 ?

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

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inverse of f(x)=2-x^2,sqrt(3)<= x<= 2

inverse

f (x) = 2 - x^{2}, 3 \leq x \leq 2

- x + 2, - - x + 2

Solution steps

y = 2 - x^{2}

Replace

x

with

y

x = 2 - y^{2}

Solve for

y, x = 2 - y^{2}

- x + 2, - - x + 2

f (x) = 3 - 29 x

f (x) = \frac{x + 9}{- 2 x - 7}

\frac{- x + 4}{- 10 x + 1}

f (x) = arctan (3)

f (x) = x^{2} - 16 x + 65

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