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

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

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

range

f (x) = \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

\frac{x ^{2} - x - 20}{x ^{2} - 4}

y = 1 x + 3, (- 7, 0)

f (x) = \frac{5 x + 1}{x - 2}

f (x) = \frac{- 2 x ^{2} + 4 x - 5}{x + 3}

f (x) = x^{3} - 12

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