Math Problem Statement

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.

Solution

The problem asks to find the slope predictor function for f(x)=x12x1/3f(x) = x - \frac{1}{2}x^{1/3} and then use it to estimate the slope of the tangent line at x=16x = 16.

Step 1: Find the slope predictor function (derivative)

To find the slope predictor function, we need to differentiate f(x)f(x) with respect to xx:

f(x)=x12x1/3f(x) = x - \frac{1}{2}x^{1/3}

Differentiate term by term:

  1. The derivative of xx is 11.

  2. The derivative of 12x1/3-\frac{1}{2}x^{1/3} requires using the power rule. Recall that ddx(xn)=nxn1\frac{d}{dx} \left( x^n \right) = n \cdot x^{n-1}.

    ddx(12x1/3)=1213x23=16x23\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)=116x23f'(x) = 1 - \frac{1}{6}x^{-\frac{2}{3}}

Step 2: Estimate the slope of the tangent line at x=16x = 16

Now, we need to plug x=16x = 16 into the slope predictor function to find the slope of the tangent line at that point:

f(16)=1161623f'(16) = 1 - \frac{1}{6} \cdot 16^{-\frac{2}{3}}

First, calculate 162316^{-\frac{2}{3}}:

  • 161/316^{1/3} is 2.51984212.5198421 (approx).
  • Then 162316^{-\frac{2}{3}} is 1(161/3)212.5198421216.34960.1574\frac{1}{(16^{1/3})^2} \approx \frac{1}{2.5198421^2} \approx \frac{1}{6.3496} \approx 0.1574.

So,

f(16)=1160.157410.02620.9738f'(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=16x = 16 is approximately 0.97380.9738.

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


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+12x1/3f(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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Slope of Tangent Line

Formulas

Derivative formula: f'(x) = lim (h -> 0) [(f(x + h) - f(x)) / h]
Power rule: d/dx (x^n) = n * x^(n-1)

Theorems

Power Rule for Differentiation

Suitable Grade Level

Grades 11-12