divergence test sum from n=1 to infinity of 1/(1+2^{1/n)}

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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

simplify solve for inverse tangent line

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Steps Examples

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average rate of change f(x)=5xln(x), 1≤ x≤ e²

Solution

10e²e²−1

Decimal

11.56517…

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Solution steps

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Average rate of change definition

The average rate of change of a function f(x) on an interval [a,b] is given by
A(x)=f(b)−f(a)b−a

=f(e²)−f(1)e²−1

f(1)=0

f(e²)=10e²

=10e²−0e²−1

10e²−0=10e²

=10e²e²−1

The average rate of change is:

=10e²e²−1

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Notebook

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

divergence test sum from n=1 to infinity of 1/(1+2^{1/n)}

Solution

Solution steps

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