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?

Question

Ezra Adams · Accepted Answer

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

Removable ones: #x#-values that make numerator and denominator 0 simultaneously.

Any function #f(x)# will be discontinuous at #x#-values that make the function undefined. Here, that will simply be any #x#-value that creates a "division by zero". What values of #x# create division by zero? In this case, it will be any #x# that satisfies #x^2-9=0# which we can factor to get #(x+3)(x-3)=0# So division by zero occurs when #x=+-3#. To find out if either of these values is a removable discontinuity, we check them one at a time to see if they also make the numerator equal to zero, thus creating a #0/0# situation. For this function, it is easy to see that when #x=3#, we get #0/0#, because #x-3# is a factor of both sides of the fraction. Thus, #x=3# is a removable discontinuity. graph{(y-(x-3)/(x^2-9))((x-3)^2+(y-0.167)^2-.02)=0 [-11.5, 8.5, -4.046, 5.95]}

Ezekiel Alexander · Answer

undefined 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.