What is the inverse of 1/(sqrt(x^2-x+1)) ?

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

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inverse of 1/(sqrt(x^2-x+1))

inverse

\frac{1}{x ^{2} - x + 1}

\frac{x + - 3 x ^{2} + 4}{2 x}, \frac{x - - 3 x ^{2} + 4}{2 x}

Solution steps

y = \frac{1}{x ^{2} - x + 1}

Replace

x

with

y

x = \frac{1}{y ^{2} - y + 1}

Solve for

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

\frac{x + - 3 x ^{2} + 4}{2 x}, \frac{x - - 3 x ^{2} + 4}{2 x}

f (x) = 6 e^{x} - 3

x + 1 - 7

g (x) = 1 \frac{2 - x}{3}

f (x) = \frac{x - 7}{x - 4}

\frac{d}{d} y = arcsin (4 x)

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