What is the inverse of 1-x^2,-1<= x<= 1 ?

The inverse of 1-x^2,-1<= x<= 1 is sqrt(-x+1),-sqrt(-x+1)

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inverse of 1-x^2,-1<= x<= 1

inverse

1 - x^{2}, - 1 \leq x \leq 1

- x + 1, - - x + 1

Solution steps

y = 1 - x^{2}

Replace

x

with

y

x = 1 - y^{2}

Solve for

y, x = 1 - y^{2}

- x + 1, - - x + 1

\frac{1}{( z - 1 ) ^{3}}

cos (2 θ)

f (x) = 1 0^{x} - 2^{3}

\frac{1}{s ^{3} + 11}

y = lo g_{2} (- x + 3)

What is the inverse of 1-x^2,-1<= x<= 1 ?
The inverse of 1-x^2,-1<= x<= 1 is sqrt(-x+1),-sqrt(-x+1)