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
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.
To determine the continuity of the function f(x) = sin(x)/(x^2 - 9), we need to consider three parts of the function separately: sin(x), x^2 - 9, and their division.
1. sin(x) is a trigonometric function that is continuous everywhere. Therefore, it is continuous for all values of x.
2. x^2 - 9 is a polynomial function. Polynomial functions are continuous for all values of x because they are composed of terms involving only the basic arithmetic operations (addition, subtraction, multiplication, and exponentiation). Thus, x^2 - 9 is continuous for all values of x.
3. The division of two continuous functions is also continuous as long as the denominator is not equal to zero. In this case, x^2 - 9 is the denominator, and it should not be equal to zero for the function to be continuous.
If we solve x^2 - 9 = 0, we get x = ±3. Therefore, the function f(x) = sin(x)/(x^2 - 9) is discontinuous at x = -3 and x = 3 since the denominator becomes zero at these points.
To summarize, the function f(x) = sin(x)/(x^2 - 9) is continuous everywhere except at x = -3 and x = 3, where it is discontinuous because the denominator becomes zero.