Understanding Continuity of Piecewise Functions

The question asks us to find the value of a that makes the given piecewise function f(x) continuous at the point x = 3.

A function f(x) is continuous at a point x = c if three conditions are met:

1. The function is defined at x = c, i.e., f(c) exists.
2. The limit of the function as x approaches c exists, i.e., \(\lim_{x \to c} f(x)\) exists.
3. The limit equals the function value, i.e., \(\lim_{x \to c} f(x) = f(c)\).

For a piecewise function, the existence of the limit \(\lim_{x \to c} f(x)\) requires that the limit from the left (\(\lim_{x \to c^-} f(x)\)) must equal the limit from the right (\(\lim_{x \to c^+} f(x)\)). Therefore, the core condition we need to satisfy for continuity at x = 3 is:

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

Calculating Limits for Continuity at x = 3

The function is defined as:

Left-Hand Limit

To find the limit as x approaches 3 from the left side (values less than 3), we use the first piece of the function, f(x) = x^2 - 1:

$$ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x^2 - 1) $$

Now, we substitute x = 3 into this expression:

$$ (3)^2 - 1 = 9 - 1 = 8 $$

So, the left-hand limit is 8.

Right-Hand Limit and Function Value

To find the limit as x approaches 3 from the right side (values greater than or equal to 3), and the function value at x = 3, we use the second piece of the function, f(x) = 2ax:

$$ \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (2ax) $$

Substituting x = 3:

$$ 2a(3) = 6a $$

And the function value at x = 3 is:

$$ f(3) = 2a(3) = 6a $$

The right-hand limit is 6a, and the function value is also 6a.

Solving for 'a'

For the function to be continuous at x = 3, the left-hand limit must equal the right-hand limit (and the function value):

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

$$ 8 = 6a $$

Now, we solve this equation for a:

$$ a = \frac{8}{6} $$

Simplifying the fraction:

$$ a = \frac{4}{3} $$

Conclusion

The value of a that makes the function f(x) continuous at x = 3 is 4/3.