What is the inverse of f(x)=8x^2+9,x>= 0 ?

The inverse of f(x)=8x^2+9,x>= 0 is (sqrt(2)sqrt(x-9))/4 ,-(sqrt(2)sqrt(x-9))/4

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inverse of f(x)=8x^2+9,x>= 0

inverse

f (x) = 8 x^{2} + 9, x \geq 0

\frac{2 x - 9}{4}, - \frac{2 x - 9}{4}

Solution steps

y = 8 x^{2} + 9

Replace

x

with

y

x = 8 y^{2} + 9

Solve for

y, x = 8 y^{2} + 9

\frac{2 x - 9}{4}, - \frac{2 x - 9}{4}

f (x) = lo g_{2} (3 x^{2} + 4)

f (x) = 3 13 x - 15

\frac{s}{s ^{2} - 1}

5 x + 20

f (x) = 5 - \frac{x}{3 x - 1}

What is the inverse of f(x)=8x^2+9,x>= 0 ?
The inverse of f(x)=8x^2+9,x>= 0 is (sqrt(2)sqrt(x-9))/4 ,-(sqrt(2)sqrt(x-9))/4