f(x)= sin x/(x^2 - 9)
Your answer is correct, but unfortunately incomplete.
In the given case, the denominator (x²-9) equals zero at x=±3, precisely what your teacher referred to.
When the denominator approaches zero, the function becomes infinite, and hence discontinuous.
Discuss the continuity of
f(x)= sin x/x^2 - 9
That is, state where is it continuous/discontinuous and why.
Here was my original answer.
sinx is cont. because it's a trig function. It's cont. at -2pi, pi, -pi, and 2pi.
x^2 is cont. because it's raised to a rational power.
-9 is cont. because it's a polynomial.
He then put x= -3 + 3? at the bottom of my work.
I'm guessing that's where it's discontinuous. So f(x) is discontinuous where x= -3 and 3
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