How to find the domain and range of the function A(x)=4x^2+1 without graphing?

Question

jdoe0001 · Answer

well
the domain is simple, $\bf \mathbb{R}$

now the range
keep in mind that, for any "x" you throw in, it'll be squared, that means including any negative values for "x", -25, -1000 and so on
once squared, it will never give you a negative number, so we can say "y" will never be negative

the lowest values $\bf x^2$ can give you is 0, and when that happens $\bf 4(0)^2+1 \implies 1$
that's the lowest "y" will get, and then it goes right back up

anonymous · Answer

so, is the domain $$(1,\infty)$$

jdoe0001 · Answer

well, the domain will be $\bf \mathbb{R} \implies (-\infty, +\infty)$

anonymous · Answer

okay, will the range be (1, infinty)

jdoe0001 · Answer

yes, the domain is not constrained by anything, the range is

anonymous · Answer

okay that makes sense. thankyou so much.

jdoe0001 · Answer

yw