Determine the domain, range, and discontinuities of the function h(x) = { 2/(x-1) if x < 3; (2/3)x - 4 if x ≥ 3 }

Steps to Solve

1. Determine the domain of ( h(x) )

To find the domain, we need to consider the conditions for both pieces of the piecewise function.

• For ( x < 3 ): The expression ( \frac{2}{x-1} ) is defined for all ( x ) except ( x = 1 ).
• For ( x \geq 3 ): The expression ( \frac{2}{3}x - 4 ) is a linear function and is defined for all ( x ).

Thus, the domain is: $$ \text{Domain} = (-\infty, 1) \cup (1, \infty) $$

2. Determine the range of ( h(x) )

Next, we will evaluate the outputs (range) from both parts of the function.

• For ( x < 3 ): The expression ( \frac{2}{x-1} ) can take any real number except it approaches infinity as ( x ) approaches 1 from the left and tends to 0 as ( x ) approaches negative infinity. Thus, the output values are ( (-\infty, 0) ) and ( (0, \infty) ).
• For ( x \geq 3 ): The linear function's output starts from its value at ( x = 3 ): $$ h(3) = \frac{2}{3}(3) - 4 = 2 - 4 = -2 $$ As ( x \to \infty ), ( h(x) ) will also approach infinity. Thus, the range is: $$ \text{Range} = (-2, \infty) $$

3. Identify discontinuities in ( h(x) )

Discontinuities occur where the function is not continuous.

• The first piece is discontinuous at ( x = 1 ), where the function becomes undefined.
• There might also be a discontinuity at ( x = 3 ) since the left hand limit from the first piece must equal the value from the second piece: $$ \lim_{x \to 3^-} h(x) = \frac{2}{2} = 1 $$ $$ h(3) = -2 $$ Since these values are not equal, there is a jump discontinuity at ( x = 3 ).

Thus, the discontinuities are: $$ \text{Discontinuities} = {1, 3} $$