# How do you find the x values at which #f(x)=(x-3)/(x^2-9)# is not continuous, which of the discontinuities are removable?

All discontinuities: set the denominator equal to 0 and solve for

Removable ones:

which we can factor to get

graph{(y-(x-3)/(x^2-9))((x-3)^2+(y-0.167)^2-.02)=0 [-11.5, 8.5, -4.046, 5.95]}

By signing up, you agree to our Terms of Service and Privacy Policy

To find the x values at which f(x) = (x-3)/(x^2-9) is not continuous, we need to identify the points where the function is undefined or where the limit does not exist.

First, we look for values of x that make the denominator, x^2-9, equal to zero. Solving x^2-9=0, we find x=3 and x=-3. These are potential points of discontinuity.

Next, we check if these points are removable discontinuities by evaluating the limit of f(x) as x approaches each of these points.

For x=3, we substitute x=3 into the function and find f(3) = (3-3)/(3^2-9) = 0/0, which is an indeterminate form. To determine if it is removable, we simplify the function by factoring the numerator and denominator. We get f(x) = (x-3)/[(x-3)(x+3)]. Canceling out the common factor (x-3), we obtain f(x) = 1/(x+3). Now, we can substitute x=3 into this simplified function and find f(3) = 1/(3+3) = 1/6. Since the limit exists and is finite, the discontinuity at x=3 is removable.

For x=-3, we substitute x=-3 into the function and find f(-3) = (-3-3)/((-3)^2-9) = -6/0, which is an indeterminate form. Simplifying the function, we have f(x) = (x-3)/[(x-3)(x+3)] = 1/(x+3). Substituting x=-3 into this simplified function, we find f(-3) = 1/0, which is undefined. Therefore, the discontinuity at x=-3 is not removable.

In summary, the function f(x) = (x-3)/(x^2-9) is not continuous at x=3 and x=-3. The discontinuity at x=3 is removable, while the discontinuity at x=-3 is not removable.

By signing up, you agree to our Terms of Service and Privacy Policy

- How do you prove that the limit of #x^2 = 9# as x approaches 3 using the epsilon delta proof?
- How do you find the limits of #lim root(3)t +12t-2t^2 where t = -oo#?
- How do you find the limit #(5+x^-1)/(1+2x^-1)# as #x->oo#?
- If limit of #f(x)=3/2# and #g(x)=1/2# as #x->c#, what the limit of #f(x)g(x)# as #x->c#?
- Lim (X^3+8/x+2). ? X=-2

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7