What is the inverse of f(x)=5-2x^2,x<= 0 ?

The inverse of f(x)=5-2x^2,x<= 0 is sqrt(-(x-5)/2),-sqrt(-(x-5)/2)

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inverse of f(x)=5-2x^2,x<= 0

inverse

f (x) = 5 - 2 x^{2}, x \leq 0

- \frac{x - 5}{2}, - - \frac{x - 5}{2}

Solution steps

y = 5 - 2 x^{2}

Replace

x

with

y

x = 5 - 2 y^{2}

Solve for

y, x = 5 - 2 y^{2}

- \frac{x - 5}{2}, - - \frac{x - 5}{2}

f (x) = - \frac{x}{( 16 - x ^{2} ) ^{\frac{1}{2}}}

(ln (x))^{6}

f (x) = 15 x + 4

f (x) = 15 x + 9

f (x) = - 0.1666 + 4.333

What is the inverse of f(x)=5-2x^2,x<= 0 ?
The inverse of f(x)=5-2x^2,x<= 0 is sqrt(-(x-5)/2),-sqrt(-(x-5)/2)