What is the range of f(x)= 6/(x^2+1) ?

The range of f(x)= 6/(x^2+1) is 0<f(x)<= 6

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range of f(x)= 6/(x^2+1)

range

f (x) = \frac{6}{x ^{2} + 1}

0 < f (x) \leq 6

Interval Notation

(0, 6]

Solution steps

Rewrite as

\frac{6}{x ^{2} + 1} = y

Multiply both sides by

x^{2} + 1

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

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

Discriminant

6 = y (x^{2} + 1) : - 4 y^{2} + 24 y

- 4 y^{2} + 24 y \geq 0 : 0 \leq y \leq 6

Check if the range interval endpoints are included

y = 0 :

Excluded

y = 6 :

Included

Therefore the range is

0 < f (x) \leq 6

- 3 x + y = 1

(- 6, 2), (4, 1)

f (x) = 3 sin (x)

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

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

What is the range of f(x)= 6/(x^2+1) ?
The range of f(x)= 6/(x^2+1) is 0<f(x)<= 6