Question

Consider the function defined by f(x)=⎩⎨⎧​2/x2 x−2 1/2(x−6)² if x<2 if 2≤x≤4 if x>4​ Graph the function. Find all values of x where the function is discontinuous. Justify your answer by addressing all the conditions for continuity.

Answer 1

The function
`\( f(x) = \begin{cases} (2)/(x^2) & \text{if } x < 2 \\ x - 2 & \text{if } 2 \leq x \leq 4 \\ (1)/(2)(x - 6)^2 & \text{if } x > 4 \end{cases} \) is discontinuous at \( x = 2 \) and \( x = 4 \).`

Step-by-step explanation:

The function provided consists of three separate expressions based on different intervals of ( x ). For continuity, three conditions need to be met: existence of the function at the point, existence of the limit at that point, and equality of the function and the limit at that point. Looking at the given function:

1. For
`\( x < 2 \)`: The expression
`\( f(x) = (2)/(x^2) \)` is a rational function. It is continuous for al
`l \( x \) except at \( x = 0 \)`(where it's undefined). Hence, it is discontinuous at ( x = 2 ) because the function changes abruptly at this point.

2. For
`\( 2 \leq x \leq 4 \):` The expression
`\( f(x) = x - 2 \)`is a linear function. It's continuous for all ( x ) in the given interval, meeting the conditions for continuity within that range.

3. For
`\( x > 4 \)`: The expression
`\( f(x) = (1)/(2)(x - 6)^2 \)` represents a quadratic function, which is continuous for all ( x ). However, at
`\( x = 4 \)`, there's a change from the linear function to the quadratic one, causing a discontinuity due to the abrupt change.

In conclusion, the function is discontinuous at
`\( x = 2 \) and \( x = 4 \)`because it fails to meet the conditions of continuity, specifically the requirement of smooth transitions between different parts of the function.

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