phạm vi ln(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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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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inverse 2log₁₀(x)

Solution

10^x2

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Find vertex using polynomial form

Find vertex using parabola form

Find vertex using vertex form

Find vertex using averaging the zeros

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3x²+2x+5y−6=0

Parabola equation in polynomial form

The vertex of an up-down facing parabola of the form y=ax²+bx+c is x_v=−b2a

Rewrite 3x²+2x+5y−6=0 in the form y=ax²+bx+c

y=−3x²5 −2x5 +65

The parabola parameters are:

a=−35 , b=−25 , c=65

x_v=−b2a

x_v=−(−25 )2(−35 )

Simplify −−25 2(−35 ) : −13

x_v=−13

Plug in x_v=−13 to find the y_v value

y_v=1915

Therefore the parabola vertex is

(−13 , 1915 )

If a<0, then the vertex is a maximum value
If a>0, then the vertex is a minimum value
a=−35

Maximum (−13 , 1915 )

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domain 2log₁₀(x)

range 2log₁₀(x)

intercepts 2log₁₀(x)

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Solve by: Find vertex using parabola form

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

phạm vi ln(x)

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