The function `f(x)= {(2 ax ", " x le 3 ),( 3x +1 ", " x gt 3):}` continuous at x= 3, then a =?

Options

A. `5/3`
B. `5/2`
C. 5
D. None of these

Answer

To find the value of \( a \) such that the function

\[ f(x) = \begin{cases} 2ax & \text{if } x \leq 3 \\ 3x + 1 & \text{if } x > 3 \end{cases} \]

is continuous at \( x = 3 \), we need to ensure that the left-hand limit, right-hand limit, and the value of the function at that point are equal.

### Step 1: Find \( f(3) \)

Since \( x = 3 \) falls under the first case of the piecewise function, we can calculate:

\[ f(3) = 2a(3) = 6a \]

### Step 2: Find the left-hand limit as \( x \) approaches 3

The left-hand limit is given by:

\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} 2ax = 2a(3) = 6a \]

### Step 3: Find the right-hand limit as \( x \) approaches 3

The right-hand limit is given by:

\[ \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (3x + 1) = 3(3) + 1 = 9 + 1 = 10 \]

### Step 4: Set the limits equal for continuity

For the function to be continuous at \( x = 3 \), we need:

\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3) \]

This gives us the equation:

\[ 6a = 10 \]

### Step 5: Solve for \( a \)

Now we can solve for \( a \):

\[ a = \frac{10}{6} = \frac{5}{3} \]

### Conclusion

Thus, the value of \( a \) for which the function is continuous at \( x = 3 \) is

\[ \boxed{\frac{5}{3}} \]

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