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

The range of 1/((x+1)^2) is f(x)>0

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range of 1/((x+1)^2)

range

\frac{1}{( x + 1 ) ^{2}}

f (x) > 0

Interval Notation

(0, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

(x + 1)^{2}

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

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

Discriminant

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

4 y \geq 0 : y \geq 0

Check if the range interval endpoints are included

y = 0 :

Excluded

Therefore the range is

f (x) > 0

f (x) = \frac{4 x ^{2} - 12 x}{x ^{2} - 9}

\frac{x}{1 - cos ( x )}

y = - 2 sin (\frac{2 π}{7} x)

h (t) = 160 t - 16 t^{2}

(0.2) (4.6)