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inverse of 1/(9+x^2)

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range x²+1x²−x−6

Solution

f(x)≤−2√2+25 or f(x)≥2√2+25

Interval Notation

(−∞ , −2√2+25 ]∪ [2√2+25 , ∞ )

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Solution steps

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Function range definition

The set of values of the dependent variable for which a function is defined

Rewrite as

x²+1x²−x−6 =y

Multiply both sides by x²−x−6

x²+1=y(x²−x−6)

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

Discriminant x²+1=y(x²−x−6): 25y²−20y−4

25y²−20y−4≥ 0 : y≤ −2√2+25 or y≥ 2√2+25

Check if the range interval endpoints are included

y=−2√2+25 : Included

y=2√2+25 : Included

Therefore the range is

f(x)≤ −2√2+25 or f(x)≥ 2√2+25

Solve instead

domain x²+1x²−x−6

inverse x²+1x²−x−6

intercepts x²+1x²−x−6

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

inverse of 1/(9+x^2)

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