What is the range of (10)/(36-x^2) ?

The range of (10)/(36-x^2) is f(x) = 5/18

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range of (10)/(36-x^2)

range

\frac{10}{36 - x ^{2}}

f (x) < 0 or f (x) \geq \frac{5}{18}

Interval Notation

(- \infty, 0) \cup [\frac{5}{18}, \infty)

Solution steps

Rewrite as

\frac{10}{36 - x ^{2}} = y

Multiply both sides by

36 - x^{2}

10 = y (36 - x^{2})

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

Discriminant

10 = y (36 - x^{2}) : 4 y (36 y - 10)

4 y (36 y - 10) \geq 0 : y \leq 0 or y \geq \frac{5}{18}

Check if the range interval endpoints are included

y = 0 :

Excluded

y = \frac{5}{18} :

Included

Therefore the range is

f (x) < 0 or f (x) \geq \frac{5}{18}

y = 4 x - 1

f (x) = \frac{3 x + 4}{5}

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

f (x) = (3 - x) e^{- x}

f (x) = (x - 1)^{2}, 0 \leq x \leq 4

What is the range of (10)/(36-x^2) ?
The range of (10)/(36-x^2) is f(x)<0\lor f(x)>= 5/18