
Anjela R.
asked 09/17/21Please help ????????
Determine the domain of the function f(x)=4/7x^2−5.
(Use symbolic notation and fractions where needed. Give your answers as intervals in the form (*,*). Use inf for infinity, U for combining intervals, and appropriate type of parenthesis"(", ")", "[" or "]" depending on whether the interval is open or closed.)
Domain is
1 Expert Answer

Raphael K. answered 09/17/21
Engineer with 10+ years of tutoring math and science
Please help ????????
Determine the domain of the function f(x)=4/7x^2−5.
(Use symbolic notation and fractions where needed. Give your answers as intervals in the form (*,*). Use inf for infinity, U for combining intervals, and appropriate type of parenthesis"(", ")", "[" or "]" depending on whether the interval is open or closed.)
Domain is all reals
For ALL polynomials its is always all reals
Domain ( -∞ , ∞ )

Raphael K.
Or (- inf, inf )09/17/21

Mark M.
What if the f(x) = [4 / (7x^2) - 5? Or f(x) = 4 / (7x^2 - 5)? The function in the problem has ambigous grouping.09/17/21

Raphael K.
That would be different. That is a rational function. 4/(7x^2 - 5) which means you need to be careful no to divide by ZERO. So that means take the denominator of the the function and set it = to 0. Like this: 7x^2 - 5 = 0 X = +/- sqrt (5/7) That tells us that there are vertical asymptotes at x = +/- sqrt (5/7) Test points before - sqrt (5/7), between +/- sqrt (5/7), and after +sqrt (5/7) to get the y values and sketch the general orientation of the curve. f(-2) = 4/23 f(-1) = 2 f(0) = -4/5 f(1) = 2 f(2) = 4/23 This implies the graph approaches 0 as x approaches - infinity. Then as the graph approaches the first asymptote at x = - sqrt (5/7), the curve goes up to + infinity, then from - infinity after -sqrt (5/7) the graph reaches -4/5 at x=0 then goes back down to - infinity as it approaches the next asymptote at + sqrt (5/7), then coming from x = +sqrt (5/7) the graph goes from -infinity to 0 as x approaches +infinity.09/17/21

Raphael K.
Domain is (-inf, -sqrt(5/7)) U (-sqrt(5/7), +sqrt(5/7)) U (+sqrt(5/7), +inf)09/17/21
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Robert S.
09/17/21