What is the range of x/(x^2-6x+8) ?

The range of x/(x^2-6x+8) is f(x) = sqrt(2)-3/2

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range of x/(x^2-6x+8)

range

\frac{x}{x ^{2} - 6 x + 8}

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

Interval Notation

(- \infty, - 2 - \frac{3}{2}] \cup [2 - \frac{3}{2}, \infty)

Solution steps

Rewrite as

\frac{x}{x ^{2} - 6 x + 8} = y

Multiply both sides by

x^{2} - 6 x + 8

x = y (x^{2} - 6 x + 8)

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

Discriminant

x = y (x^{2} - 6 x + 8) : 4 y^{2} + 12 y + 1

4 y^{2} + 12 y + 1 \geq 0 : y \leq - 2 - \frac{3}{2} or y \geq 2 - \frac{3}{2}

Check if the range interval endpoints are included

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

Included

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

Included

Therefore the range is

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

f (x) = \frac{( 3 x - 1 )}{( 8 x - 3 )}

f (x) = 15 x^{4} - 15 x^{2}

f (x) = \frac{2 x + 5}{x - 7}

f (x) = \frac{2 x - 8}{3 x ^{2} - 7 x - 20}

f (x) = \frac{3 x + 4}{x ^{2} + x - 6}

What is the range of x/(x^2-6x+8) ?
The range of x/(x^2-6x+8) is f(x)<=-sqrt(2)-3/2 \lor f(x)>= sqrt(2)-3/2