What is the range of f(x)=(4x^2-5)/(2x^2+8) ?

The range of f(x)=(4x^2-5)/(2x^2+8) is -5/8 <= f(x)<2

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range of f(x)=(4x^2-5)/(2x^2+8)

range

f (x) = \frac{4 x ^{2} - 5}{2 x ^{2} + 8}

- \frac{5}{8} \leq f (x) < 2

Interval Notation

[- \frac{5}{8}, 2)

Solution steps

Rewrite as

\frac{4 x ^{2} - 5}{2 x ^{2} + 8} = y

Multiply both sides by

2 x^{2} + 8

4 x^{2} - 5 = y (2 x^{2} + 8)

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

Discriminant

4 x^{2} - 5 = y (2 x^{2} + 8) : - 64 y^{2} + 88 y + 80

- 64 y^{2} + 88 y + 80 \geq 0 : - \frac{5}{8} \leq y \leq 2

Check if the range interval endpoints are included

y = - \frac{5}{8} :

Included

y = 2 :

Excluded

Therefore the range is

- \frac{5}{8} \leq f (x) < 2

- tan (x)

4 x^{2} - 14 x + 8

f (x) = x - 2

f (x) = 6 x^{4} + 24 x^{3}

f (x) = 1 + e^{- x}

What is the range of f(x)=(4x^2-5)/(2x^2+8) ?
The range of f(x)=(4x^2-5)/(2x^2+8) is -5/8 <= f(x)<2