Domain and Range of f(x,y)=2x^2+3x^2+1: Real Numbers and Minimum Value of 1

[seto6]

Homework Statement

Find the domain and the range

f(x,y)=2x²+3x²+1

Homework Equations

non

The Attempt at a Solution

...
i said that x,y are all real #'s

and f(x,y)>=1 or z>=1

is there anything wrong with what i has attempted at?

if do could you explain...

thanks in advance

 

[fluxions]

check out (x,y) = (-1,1).

 

[seto6]

what do you mean by that?

 

[HallsofIvy]

I assume you mean f(x,y)=2x^2+3y^2+1

First, the "natural" domain for a given formula is the set of all x values for which the formula can be calculated. There is no reason why we cannot square any number or multiply or add any numbers. The domain for this function, like the domain for any polynomial is "all real numbers" for both x and y or, more formally, \mathtype{R}^2.

A far as the range is concerned, we know that a square is never negative so neither 2x^2 nor 3y^2 can be less than 0. That means that f(x,y)= 2x^2+ 3y^2+ 1 is never less than 1 but can, of course, have any value above that.

(I have no idea what fluxions meant.)

 

[fluxions]

The OP edited the function after my post. It was originally f(x,y) = 2x^2 + 3x^y + 1, and s/he made the claim that f(x,y) > 1 for all x,y. I suggested that was false, a counterexample being f(-1,1). Sorry for the confusion.