What is the inverse of f(x)=9-2x^2,x<= 0 ?

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

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inverse of f(x)=9-2x^2,x<= 0

inverse

f (x) = 9 - 2 x^{2}, x \leq 0

- \frac{x - 9}{2}, - - \frac{x - 9}{2}

Solution steps

y = 9 - 2 x^{2}

Replace

x

with

y

x = 9 - 2 y^{2}

Solve for

y, x = 9 - 2 y^{2}

- \frac{x - 9}{2}, - - \frac{x - 9}{2}

f (x) = lo g_{2} (4 x + 7)

f (x) = \frac{- 2 x + 5}{- 3 x + 9}

y = \frac{2 x + 2}{5 x - 3}

f (x) = arctan (x), 0 < x \leq 1

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

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