Math, asked by superdragonxy, 8 months ago

what is the domain and range of 5/(4-x^2)​

Answers

Answered by mitronmitron364
1
Domain: (−∞,−1)∪(−1,1)∪(1,∞)
Range: (−∞,−4]∪(0,∞)

Explanation:

Best explained through the graph.
graph{4/(x^2-1) [-5, 5, -10, 10]}
We can see that for the domain, the graph starts at negative infinity. It then hits a vertical asymptote at x = -1.
That's fancy math-talk for the graph is not defined at x = -1, because at that value we have 4(−1)2−1 which equals 41−1 or 40.
Since you can't divide by zero, you can't have a point at x = -1, so we keep it out of the domain (recall that the domain of a function is the collection of all the x-values that produce a y-value).
Then, between -1 and 1, everything's fine, so we have to include it in the domain.
Things start getting funky at x = 1 again. Once more, when you plug in 1 for x, the result is 40 so we have to exclude that from the domain.
To sum it up, the function's domain is from negative infinity to -1, then from -1 to 1, and then to infinity. The mathy way of expressing that is (−∞,−1)∪(−1,1)∪(1,∞).
The range follows the same idea: it's the set of all y-values of the function. We can see from the graph that from negative infinity to -4, all is well.
Then things start going south. At y=-4, x=0; but then, if you try y=-3, you won't get an x. Watch:

−3=4x2−1
−3(x2−1)=4
x2−1=−43
x2=−43+1=−13
x=√−13

There is no such thing as the square root of a negative number. That's saying some number squared equals −13, which is impossible because squaring a number always has a positive result.
That means y=-3 is undefined and so is not part of our range. The same is true for all y-values between 4 and 0.
From 0 above, everything is good all the way to infinity. Our range is then negative infinity to -4, then 0 to infinity; in math terms, (−∞,−4]

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