What is the inverse of 2x+sqrt(4+x^2),-3<x<2 ?

The inverse of 2x+sqrt(4+x^2),-3<x<2 is (2x+sqrt(x^2+12))/3 ,(2x-sqrt(x^2+12))/3

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inverse of 2x+sqrt(4+x^2),-3<x<2

inverse

2 x + 4 + x^{2}, - 3 < x < 2

\frac{2 x + x ^{2} + 12}{3}, \frac{2 x - x ^{2} + 12}{3}

Solution steps

y = 2 x + 4 + x^{2}

Replace

x

with

y

x = 2 y + 4 + y^{2}

Solve for

y, x = 2 y + 4 + y^{2}

\frac{2 x + x ^{2} + 12}{3}, \frac{2 x - x ^{2} + 12}{3}

f (x) = 3 sin (x) - 6

f (x) = 6 x^{2} + 9 x - 2

f (x) = \frac{( - y ^{2} + 6 y + 28 )}{60}

f (x) = (\frac{3}{2})

\frac{- 3 x ^{2} - 2}{2}

What is the inverse of 2x+sqrt(4+x^2),-3<x<2 ?
The inverse of 2x+sqrt(4+x^2),-3<x<2 is (2x+sqrt(x^2+12))/3 ,(2x-sqrt(x^2+12))/3