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

The range of f(x)=(-1)/(x^2-2x+1) is f(x)<0

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

range

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

f (x) < 0

Interval Notation

(- \infty, 0)

Solution steps

Rewrite as

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

Simplify

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

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

Multiply both sides by

x^{2} - 2 x + 1

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

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

Discriminant

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

- 4 y \geq 0 : y \leq 0

Check if the range interval endpoints are included

y = 0 :

Excluded

Therefore the range is

f (x) < 0

y = \frac{1}{5} (x + 4)

f (x) = lo g_{\frac{1}{4}} (x^{5})

f (x) = 2 cos (3 x) + 1

\frac{1}{6} x^{3} - 3

y = 5 x - 3

What is the range of f(x)=(-1)/(x^2-2x+1) ?
The range of f(x)=(-1)/(x^2-2x+1) is f(x)<0