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The problem asks to find the slope predictor function for \( f(x) = x - \frac{1}{2}x^{1/3} \) and then use it to estimate the slope of the tangent line at \( x = 16 \).

### Step 1: Find the slope predictor function (derivative)
To find the slope predictor function, we need to differentiate \( f(x) \) with respect to \( x \):

\[ f(x) = x - \frac{1}{2}x^{1/3} \]

Differentiate term by term:

1. The derivative of \( x \) is \( 1 \).
2. The derivative of \( -\frac{1}{2}x^{1/3} \) requires using the power rule. Recall that \( \frac{d}{dx} \left( x^n \right) = n \cdot x^{n-1} \).

   \[
   \frac{d}{dx}\left(-\frac{1}{2}x^{1/3}\right) = -\frac{1}{2} \cdot \frac{1}{3} \cdot x^{-\frac{2}{3}} = -\frac{1}{6}x^{-\frac{2}{3}}
   \]

So, the slope predictor function (derivative) is:

\[
f'(x) = 1 - \frac{1}{6}x^{-\frac{2}{3}}
\]

### Step 2: Estimate the slope of the tangent line at \( x = 16 \)
Now, we need to plug \( x = 16 \) into the slope predictor function to find the slope of the tangent line at that point:

\[
f'(16) = 1 - \frac{1}{6} \cdot 16^{-\frac{2}{3}}
\]

First, calculate \( 16^{-\frac{2}{3}} \):

- \( 16^{1/3} \) is \( 2.5198421 \) (approx).
- Then \( 16^{-\frac{2}{3}} \) is \( \frac{1}{(16^{1/3})^2} \approx \frac{1}{2.5198421^2} \approx \frac{1}{6.3496} \approx 0.1574 \).

So,

\[
f'(16) = 1 - \frac{1}{6} \cdot 0.1574 \approx 1 - 0.0262 \approx 0.9738
\]

### Final Answer
The slope of the tangent line at \( x = 16 \) is approximately \( 0.9738 \).

Would you like to see more details, or do you have any further questions?

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Here are five related questions to further explore the topic:

1. How do you interpret the slope of a tangent line in the context of a function?
2. What does it mean if the slope of the tangent line is positive or negative?
3. How would the slope predictor function change if the original function were \( f(x) = x + \frac{1}{2}x^{1/3} \)?
4. How can you use the second derivative to determine if the tangent line slope is increasing or decreasing at a point?
5. How does the slope of a tangent line relate to the concept of instantaneous rate of change?

**Tip:** Always check your derivative by simplifying and breaking down complex expressions to avoid calculation errors.

Find the slope predictor function for f(x) = x - (1/2)x^(1/3) and use it to estimate the slope of the tangent line at x = 16.

Learn how to find the slope predictor function for f(x) = x - (1/2)x^(1/3) and estimate the slope of the tangent line at x = 16. This solution uses calculus concepts such as differentiation, the power rule, and finding the slope of a tangent line. Suitable for students in Grades 11-12.

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