Solve for n. 3 (n 5) ≥ 3n 8 a. no solution b. n ≥ 23 c. n ≥ 7 d. all real numbers
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A discontinuity is a point that cannot exist because the x-coordinate would cause a problem in the equation. If you have a polynomial in the denominator, you must find which values of x would cause the polynomial in the denominator to evaluate to zero. Since division by zero is undefined, that would cause a discontinuity.
Let's look at your function.
is in the numerator. It is defined for every value of x. There is no problem there.
is in the denominator. This is a function defined for every value of x, but since it is in the denominator, we must exclude the x-value that would cause this polynomial to evaluate to zero.
We set it equal to zero and solve the equation for x.
For x = 0, the denominator has a value of zero, so at this point there is a discontinuity in function f(x).
The answer is:
Let's look at your function.
We set it equal to zero and solve the equation for x.
For x = 0, the denominator has a value of zero, so at this point there is a discontinuity in function f(x).
The answer is: