Question

Discuss the continuity of the function (x). (consider all the possible points of discontinuity. in your solution, you must present if discontinuous – determine the type of discontinuity and indicate if continuous at the left/right). g(x)={ (x^2 - x - 2)/(x 1) x<=1, sq of (5-x) 1=5 a. what are all the possible points of discontinuity of ? b. explain the continuity of at the points in letter (a). if the function is discontinuous, do not forget to determine the type of discontinuity and indicate if continuous at the left/right. in addition, if there is(are) a removable discontinuity(discontinuities), redefine the function.

Answer 1

The possible points of discontinuity for the function
`\(g(x)\) are \(x = 1\) and \(x = 5\)`. The function is discontinuous at
`\(x = 1\)` with a removable discontinuity, and it is continuous at
`\(x = 5\)`.

Step-by-step explanation:

The function
`\(g(x)\)` is defined as follows:

\[ g(x) = \begin{cases}

`(x^2 - x - 2)/(x - 1) & \text{if } x \leq 1 \\√(5 - x) & \text{if } x > 1`

`\end{cases} \]`

Now, let's analyze the possible points of discontinuity:

1. Discontinuity at
`\(x = 1\):At \(x = 1\)`, the denominator becomes zero, leading to an undefined value. To determine the type of discontinuity, factorize the numerator:
`\(x^2 - x - 2 = (x - 2)(x + 1)\)`. Canceling the common factor of
`\(x - 1\)` in both the numerator and denominator, we get
`\(g(x) = x + 2\) for \(x \leq 1\)`. So, the discontinuity
`at \(x = 1\)` is removable, and
`\(g(x)\)`can be redefined as
`\(g(x) = x + 2\) for \(x \leq 1\).`

2. Discontinuity at
`\(x = 5\):At \(x = 5\)`, the square root term
`\(√(5 - x)\)` is defined for
`\(x < 5\)`, so there is no discontinuity at
`\(x = 5\)`.

In summary, the function has a removable discontinuity at
`\(x = 1\)`and is continuous at
`\(x = 5\)`. The redefined function is
`\(g(x) = x + 2\) for \(x \leq 1\)`.