What is the range of f(x)=(-5)/(x^2+1) ?

The range of f(x)=(-5)/(x^2+1) is -5<= f(x)<0

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range of f(x)=(-5)/(x^2+1)

range

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

- 5 \leq f (x) < 0

Interval Notation

[- 5, 0)

Solution steps

Rewrite as

\frac{- 5}{x ^{2} + 1} = y

Simplify

\frac{- 5}{x ^{2} + 1} : - \frac{5}{x ^{2} + 1}

- \frac{5}{x ^{2} + 1} = y

Multiply both sides by

x^{2} + 1

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

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

Discriminant

- 5 = y (x^{2} + 1) : - 4 y^{2} - 20 y

- 4 y^{2} - 20 y \geq 0 : - 5 \leq y \leq 0

Check if the range interval endpoints are included

y = - 5 :

Included

y = 0 :

Excluded

Therefore the range is

- 5 \leq f (x) < 0

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

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

f (x) = \frac{6 e ^{x}}{e ^{x} - 3}

f (x) = c

\frac{e ^{x} - e ^{- x}}{2}

What is the range of f(x)=(-5)/(x^2+1) ?
The range of f(x)=(-5)/(x^2+1) is -5<= f(x)<0