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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

range

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

0 < f (x) \leq 5

Interval Notation

(0, 5]

Solution steps

Rewrite as

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

Multiply both sides by

2 x^{2} + 1

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

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

Discriminant

5 = y (2 x^{2} + 1) : - 8 y^{2} + 40 y

- 8 y^{2} + 40 y \geq 0 : 0 \leq y \leq 5

Check if the range interval endpoints are included

y = 0 :

Excluded

y = 5 :

Included

Therefore the range is

0 < f (x) \leq 5

(5, - 4), (- 1, - 4)

\frac{x ^{3}}{x + 2}

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

f (x) = x^{2} + x

f (x) = \frac{x - 1}{x ^{2} - 4 x + 3}

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