What is the inverse function of $y = 4 x^{2} + 3$ $y=4x^2+3$ ?

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Answer 1 · 2017-04-10T05:08:52

Inverse function is $\frac{\sqrt{x - 3}}{2}$ $sqrt(x-3)/2$

Let $f (x) = y = 4 x^{2} + 3$ $f(x)=y=4x^2+3$

Therefore $4 x^{2} = y - 3$ $4x^2=y-3$

and $x^{2} = \frac{y - 3}{4}$ $x^2=(y-3)/4$ i.e. $x = \frac{\sqrt{y - 3}}{2}$ $x=sqrt(y-3)/2$

Hence, inverse function of $f (x) = 4 x^{2} + 3$ $f(x)=4x^2+3$ is $f^{- 1} (x) = \frac{\sqrt{x - 3}}{2}$ $f^(-1)(x)=sqrt(x-3)/2$

What is the inverse function of y=4x^2+3y=4x2+3y=4x^2+3?