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4|x+10|>= 8

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range y=xx²−6x+8

Solution

f(x)≤ −√2−32 or f(x)≥√2−32

Interval Notation

(−∞ , −√2−32 ]∪ [√2−32 , ∞ )

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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

xx²−6x+8 =y

Multiply both sides by x²−6x+8

x=y(x²−6x+8)

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

Discriminant x=y(x²−6x+8): 4y²+12y+1

4y²+12y+1≥ 0 : y≤ −√2−32 or y≥ √2−32

Check if the range interval endpoints are included

y=−√2−32 : Included

y=√2−32 : Included

Therefore the range is

f(x)≤ −√2−32 or f(x)≥ √2−32

Solve instead

domain xx²−6x+8

inverse xx²−6x+8

intercepts xx²−6x+8

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

4|x+10|>= 8

Solution

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