solve for the domain and range of a) f(x)= 3-2x^2 b) h(x)= the square root of 36-x^2
"solve for" is not a good instruction. Generally... Domain: What values can you put in? a) Everything works, here. All Real Numbers b) Whoops \(36 - x^{2} \ge 0\) Do you see why?
huh? what about the range
range of \(f\) \(x^2\geq 0\) always, so \(-2x^2\leq 0\) and therefore \(3-2x^2\leq 3\)
\[h(x)=\sqrt{36-x^2}\] if you start with a circle centered at the origin with radius \(6\) \[x^2+y^2=36\]and solve for \(y\) you get \[y=\pm\sqrt{36-x^2}\]
if you get rid of the \(\pm\) and just put \(+\) you have \[h(x)=\sqrt{36-x^2}\] which is the upper half of the circle
I'll take that as a "no". Like I tried to say, this is a thought problem. You have to ponder it and try to see what it means. There isn't necessarily a beautiful equation to provide what you need. Go exploring!
ludvic: I expect it would help you a lot if you review the meaning of "domain" and "range", online, in a texbook, or following tkhunny's examples and informal definition. Another suggestion is that you try graphing both of the given equations. Then you'd have something visual to compare against the definitions of "domain" and "range."
first do you get the domain part of this question
The first half of this post involves the function f(x)= 3-2x^2. You are asked to find the domain of this function. As tkhunny has hinted, "Domain: What values can you put in?" In other words, which x values are acceptable inputs to the polynomial function f(x)= 3-2x^2? It'd be wise to ask yourself also whether there are any x-values that would NOT be acceptable inputs to f(x)= 3-2x^2. As before, I'd suggest that you actually sketch the graph of f(x)= 3-2x^2 and use the image to help you determine whether or not there are restrictions to the values of x that can serve as inputs to f(x)= 3-2x^2.
@ludvic: i invite you to become involved here. The mroe involved you are, the more you're likely to learn from this discussion.
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