phạm vi 4x^2+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

simplify solve for inverse tangent line

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range 4x²+1

Solution

f(x)≥ 1

Interval Notation

[1, ∞ )

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Function range definition

The set of values of the dependent variable for which a function is defined

Vertex of 4x²+1: Minimum (0, 1)

For a parabola ax²+bx+c with Vertex (x_v, y_v)
If a<0 the range is f(x)≤ y_v
If a>0 the range is f(x)≥ y_v
a=4, Vertex (x_v, y_v)=(0, 1)

f(x)≥ 1

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domain 4x²+1

vertices 4x²+1

inverse 4x²+1

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

phạm vi 4x^2+1

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