Let `f(x) = {(3 , ,xlt1),(4-x,,xgt1):}, ` Investigate x=1 for the existence of a local maxima/minima.

Answer

To investigate the existence of a local maxima or minima at \( x = 1 \) for the function defined as:

\[ f(x) = \begin{cases} 3 & \text{if } x < 1 \\ 4 - x & \text{if } x > 1 \end{cases} \]

we will follow these steps:

### Step 1: Determine the value of \( f(1) \)

Since \( f(x) \) is defined piecewise, we need to evaluate \( f(1) \) directly. However, \( f(1) \) is not defined in the given piecewise function. We need to check the left-hand and right-hand limits at \( x = 1 \).

\[ f(1) = \lim_{x \to 1^-} f(x) = 3 \quad (\text{as } x \text{ approaches } 1 \text{ from the left}) \]
\[ f(1) = \lim_{x \to 1^+} f(x) = 4 - 1 = 3 \quad (\text{as } x \text{ approaches } 1 \text{ from the right}) \]

Thus, we can conclude that:

\[ f(1) = 3 \]

### Step 2: Analyze the behavior of \( f(x) \) around \( x = 1 \)

Next, we will check the values of \( f(x) \) around \( x = 1 \):

- For \( x < 1 \), \( f(x) = 3 \).
- For \( x > 1 \), \( f(x) = 4 - x \).

Now, we can evaluate \( f(x) \) for values just less than and just greater than 1:

- For \( x = 0.9 \) (which is less than 1):
\[ f(0.9) = 3 \]

- For \( x = 1.1 \) (which is greater than 1):
\[ f(1.1) = 4 - 1.1 = 2.9 \]

### Step 3: Compare the values

Now we compare the values of \( f(x) \) around \( x = 1 \):

- \( f(0.9) = 3 \)
- \( f(1) = 3 \)
- \( f(1.1) = 2.9 \)

### Step 4: Determine if there is a local maxima or minima

From the values calculated, we observe:

- As \( x \) approaches 1 from the left, \( f(x) \) remains constant at 3.
- As \( x \) moves to the right of 1, \( f(x) \) decreases to 2.9.

Since \( f(x) \) is greater than or equal to \( f(1) \) for values less than 1 and less than \( f(1) \) for values greater than 1, we can conclude that:

- \( x = 1 \) is a point of local maxima.

### Final Conclusion

Thus, the function \( f(x) \) has a local maximum at \( x = 1 \) with \( f(1) = 3 \).

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