What is the inverse of f(x)= 3/4 x^2+1 ?

The inverse of f(x)= 3/4 x^2+1 is sqrt((4x-4)/3),-sqrt((4x-4)/3)

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inverse of f(x)= 3/4 x^2+1

inverse

f (x) = \frac{3}{4} x^{2} + 1

\frac{4 x - 4}{3}, - \frac{4 x - 4}{3}

Solution steps

y = \frac{3}{4} x^{2} + 1

Replace

x

with

y

x = \frac{3}{4} y^{2} + 1

Solve for

y, x = \frac{3}{4} y^{2} + 1

\frac{4 x - 4}{3}, - \frac{4 x - 4}{3}

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

f (x) = \frac{x ^{2} + 5 x - 14}{x ^{2} - 4}

f (x) = x + 3 - 2

(x + 1)^{2}

\frac{1}{x - 5}

What is the inverse of f(x)= 3/4 x^2+1 ?
The inverse of f(x)= 3/4 x^2+1 is sqrt((4x-4)/3),-sqrt((4x-4)/3)