What is the inverse of g(x)=8e^{x^2+1} ?

The inverse of g(x)=8e^{x^2+1} is sqrt(ln(x/8)-1),-sqrt(ln(x/8)-1)

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inverse of g(x)=8e^{x^2+1}

inverse

g (x) = 8 e^{x^{2} + 1}

ln (\frac{x}{8}) - 1, - ln (\frac{x}{8}) - 1

Solution steps

y = 8 e^{x^{2} + 1}

Replace

x

with

y

x = 8 e^{y^{2} + 1}

Solve for

y, x = 8 e^{y^{2} + 1}

ln (\frac{x}{8}) - 1, - ln (\frac{x}{8}) - 1

f (x) = x^{3} - 12 x^{2} - 27 x + 9

f (x) = - x^{4} - 4 x^{3} + 8 x - 1

f (x) = 2 x^{3} - 3 x^{2} + 5 x - 5

f (x) = \frac{x - 9}{4}

f (x) = 5 sin (x)

What is the inverse of g(x)=8e^{x^2+1} ?
The inverse of g(x)=8e^{x^2+1} is sqrt(ln(x/8)-1),-sqrt(ln(x/8)-1)