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

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

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range of x/(x^2-x+1)

range

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

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

Interval Notation

[- \frac{1}{3}, 1]

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} - x + 1

x = y (x^{2} - x + 1)

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

Discriminant

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

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

Check if the range interval endpoints are included

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

Included

y = 1 :

Included

Therefore the range is

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

(0.8) (- 4.4)

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

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

4 x + 11

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