inverse ofr sec^2(x)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

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log_□

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ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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range f(x)=1x²

Solution

f(x)>0

Interval Notation

(0, ∞ )

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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

1x² =y

Multiply both sides by x²

1=yx²

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

Discriminant 1=yx²: 4y

4y≥ 0 : y≥ 0

Check if the range interval endpoints are included

y=0 : Excluded

Therefore the range is

f(x)>0

Solve instead

domain 1x²

inverse 1x²

intercepts 1x²

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

inverse ofr sec^2(x)

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