taylor ln(x+1)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

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

lhopital squeeze chain factoring sandwich

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domain tan(2x−5)

Solution

π2 n≤ x<10−3π4 +π2 n or 10−3π4 +π2 n<x<π2 +π2 n

Interval Notation

[π2 n, 10−3π4 +π2 n)∪(10−3π4 +π2 n, π2 +π2 n)

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Solution steps

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Domain definition

The domain of a function is the set of input or argument values for which the function is real and defined

Take the denominator(s) of tan(2x−5) and compare to zero: x=10−3π4 +π2 n

Periodicity of tan(2x−5): π2

Combine period: π2 n≤ x<π2 +π2 n and undefined points.
The domain is

π2 n≤ x<10−3π4 +π2 n or 10−3π4 +π2 n<x<π2 +π2 n

Solve instead

range tan(2x−5)

periodicity tan(2x−5)

amplitude tan(2x−5)

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