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

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

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range of f(x)= x/(x^2-4x+3)

range

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

f (x) \leq - \frac{3}{2} - 1 or f (x) \geq \frac{3}{2} - 1

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} - 4 x + 3

x = y (x^{2} - 4 x + 3)

The range is the set of y for which the discriminant is greater or equal to zero

Discriminant

x = y (x^{2} - 4 x + 3) : 4 y^{2} + 8 y + 1

4 y^{2} + 8 y + 1 \geq 0 : y \leq - \frac{3}{2} - 1 or y \geq \frac{3}{2} - 1

Check if the range interval endpoints are included

y = - \frac{3}{2} - 1 :

Included

y = \frac{3}{2} - 1 :

Included

Therefore the range is

f (x) \leq - \frac{3}{2} - 1 or f (x) \geq \frac{3}{2} - 1

y = \frac{3 x ^{2} - 3 x}{x ^{2} + x - 12}

\frac{- 7}{2 x ^{\frac{3}{2}}}

f (x) = 9 x - 7 x^{2}

f (x) = 3 sin (\frac{6 x}{7} + 2 π)

\frac{x + 6}{x - 5}

What is the range of f(x)= x/(x^2-4x+3) ?
The range of f(x)= x/(x^2-4x+3) is f(x)<=-(sqrt(3))/2-1\lor f(x)>= (sqrt(3))/2-1