What is the range of $y=-3 \cdot 2^{\sqrt{x}}$ ?

Step 1 ：The function given is \(y=-3 \cdot 2^{\sqrt{x}}\).

Step 2 ：The base of the exponential function is 2, which is positive. The exponent is \(\sqrt{x}\), which is always non-negative for real x. Therefore, \(2^{\sqrt{x}}\) is always positive.

Step 3 ：When we multiply this by -3, the result is always negative.

Step 4 ：The limit of the function as x approaches infinity is negative infinity, and the limit as x approaches 0 is -3. This means that the function covers all values from -3 to negative infinity.

Step 5 ：\(\boxed{\text{Final Answer: The range of the function } y=-3 \cdot 2^{\sqrt{x}} \text{ is } (-\infty, -3]}\)