Math Problem Statement

The height of a roller coaster at a given time is modeled by the polynomial function h(t) = (-0.01t^3 + 0.35t^2 - 3.5t + 10)(t + 5)^2(t - 5)(t + 1)(2 - t^(1/3)), where t represents time in seconds. Find the turning point(s) of the height function.

Solution

The problem asks to find the turning points of the height function of a roller coaster, which is modeled by the following polynomial function:

h(t)=(0.01t3+0.35t23.5t+10)(t+5)2(t5)(t+1)(2t13)h(t) = \left(-0.01t^3 + 0.35t^2 - 3.5t + 10\right)(t + 5)^2(t - 5)(t + 1)(2 - t^{\frac{1}{3}})

Steps to solve:

  1. Find the derivative of h(t)h(t): The turning points of a function occur where its derivative is zero. So, the first step is to find h(t)h'(t).

  2. Set h(t)=0h'(t) = 0: Solve the derivative equation to find the values of tt where the slope is zero, which gives potential turning points.

  3. Determine the nature of turning points: Use the second derivative test to classify whether each critical point is a maximum, minimum, or inflection point.

Let me walk you through each of these steps. Would you like to focus on one step at a time or get a full solution right away?


In the meantime, consider these related questions:

  1. How do turning points relate to the behavior of the roller coaster's height?
  2. Why is the derivative zero at turning points?
  3. How does the product rule apply in the first derivative of h(t)h(t)?
  4. What effect do the factors (t+5)2(t + 5)^2 and (t5)(t - 5) have on the turning points?
  5. What does the cube root term t13t^{\frac{1}{3}} contribute to the derivative?

Tip:

For solving derivatives of products like this, the product rule combined with chain rule for terms like (2t13)(2 - t^{\frac{1}{3}}) will be essential!

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Turning Points
Polynomial Functions

Formulas

First Derivative Test
Second Derivative Test

Theorems

Product Rule
Chain Rule

Suitable Grade Level

Grades 11-12 (Advanced High School or Early College)