Question:Which of the following functions has(have) a minimum value at x = 2?\(\mathrm{f(x) = \frac{1}{2}(x-2)^2 + 5}\)\(\mathrm{g(x) = 5\left(\frac{1...

1. INFER what type of functions we're analyzing

We have two functions to examine:

• Function I: \(\mathrm{f(x) = \frac{1}{2}(x-2)^2 + 5}\)
• Function II: \(\mathrm{g(x) = 5(\frac{1}{2})^x + 2}\)

The key insight is recognizing what type of function each expression represents, as different function types have different behaviors regarding extrema.

2. INFER the properties of Function I

Function I has the form \(\mathrm{y = a(x-h)^2 + k}\), which is vertex form of a quadratic equation:

• \(\mathrm{a = \frac{1}{2}}\) (positive coefficient)
• \(\mathrm{h = 2, k = 5}\) (vertex coordinates)

Since \(\mathrm{a \gt 0}\), this parabola opens upward. Upward-opening parabolas have their minimum value at the vertex.

Therefore: \(\mathrm{f(x)}\) has a minimum value at \(\mathrm{x = 2}\), and that minimum value is \(\mathrm{f(2) = 5}\).

3. INFER the properties of Function II

Function II has the form \(\mathrm{y = ab^x + c}\) where:

• \(\mathrm{a = 5}\) (positive leading coefficient)
• \(\mathrm{b = \frac{1}{2}}\) (base between 0 and 1)
• \(\mathrm{c = 2}\) (vertical shift)

Since \(\mathrm{0 \lt b \lt 1}\), this is exponential decay. Exponential decay functions are strictly decreasing everywhere—they never "turn around" to create local minima or maxima.

4. INFER the final answer

• Function I: Has minimum at \(\mathrm{x = 2}\) ✓
• Function II: Strictly decreasing, no minimum anywhere ✗

Only Function I has a minimum at \(\mathrm{x = 2}\).

Answer: A. I only

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Not recognizing that exponential decay functions are strictly decreasing and therefore cannot have minimum values.

Students might think: "Let me check if \(\mathrm{g(2)}\) gives a smaller value than nearby points" and start calculating \(\mathrm{g(1), g(2), g(3)}\) without realizing that since \(\mathrm{g(x)}\) decreases continuously, \(\mathrm{g(x)}\) will always be smaller for larger x values. They may conclude that \(\mathrm{g(x)}\) has a minimum at \(\mathrm{x = 2}\) simply because they didn't check enough points or understand the function's global behavior.

This may lead them to select Choice C (I and II).

Second Most Common Error:

Missing conceptual knowledge: Not remembering that vertex form \(\mathrm{y = a(x-h)^2 + k}\) has its vertex at \(\mathrm{(h,k)}\), or forgetting that the sign of 'a' determines whether it's a minimum or maximum.

Students might misidentify the vertex location or think that since there's a squared term, it automatically means there's a minimum without checking the coefficient sign. This creates confusion about Function I's behavior.

This leads to confusion and guessing between the remaining choices.

The Bottom Line:

This problem requires students to quickly identify function types and connect those types to their extrema behavior. The key insight is that different function families have fundamentally different behaviors—quadratics can have extrema, but exponential functions (except constant ones) do not.