Math Problem Statement

Find the absolute maximum and absolute minimum values of f(t) = t - ³√t on the interval [-1, 6]

Solution

To solve for the absolute maximum and absolute minimum of the function f(t)=tt3f(t) = t - \sqrt[3]{t} on the interval [1,6][-1, 6], we will follow these steps:

Step 1: Find the derivative of the function

First, we need to find the critical points by taking the derivative of f(t)=tt3f(t) = t - \sqrt[3]{t}.

f(t)=tt1/3f(t) = t - t^{1/3}

The derivative is:

f(t)=113t2/3f'(t) = 1 - \frac{1}{3}t^{-2/3}

Simplifying:

f(t)=113t23f'(t) = 1 - \frac{1}{3\sqrt[3]{t^2}}

Step 2: Set the derivative equal to 0 to find critical points

We solve f(t)=0f'(t) = 0:

113t23=01 - \frac{1}{3\sqrt[3]{t^2}} = 0

13t23=1\frac{1}{3\sqrt[3]{t^2}} = 1

t23=13\sqrt[3]{t^2} = \frac{1}{3}

Cubing both sides:

t2=(13)3=127t^2 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}

t=±1273=±13t = \pm \frac{1}{\sqrt[3]{27}} = \pm \frac{1}{3}

So, the critical points are t=13t = \frac{1}{3} and t=13t = -\frac{1}{3}.

Step 3: Evaluate the function at critical points and endpoints

We now evaluate f(t)f(t) at the critical points and the endpoints t=1t = -1 and t=6t = 6.

  • f(1)=113=1+1=0f(-1) = -1 - \sqrt[3]{-1} = -1 + 1 = 0
  • f(6)=66361.8174.183f(6) = 6 - \sqrt[3]{6} \approx 6 - 1.817 \approx 4.183
  • f(13)=131330.3330.6930.36f\left(\frac{1}{3}\right) = \frac{1}{3} - \sqrt[3]{\frac{1}{3}} \approx 0.333 - 0.693 \approx -0.36
  • f(13)=13133=13+0.6930.36f\left(-\frac{1}{3}\right) = -\frac{1}{3} - \sqrt[3]{-\frac{1}{3}} = -\frac{1}{3} + 0.693 \approx 0.36

Step 4: Determine absolute minimum and maximum

  • The minimum value is f(13)0.36f\left(\frac{1}{3}\right) \approx -0.36.
  • The maximum value is f(6)4.183f(6) \approx 4.183.

Thus:

  • The absolute minimum is approximately 0.36-0.36.
  • The absolute maximum is approximately 4.1834.183.

Let me know if you need further clarification or additional details!

Follow-up questions:

  1. What are the exact critical points of a function with a cubic root?
  2. Why are the endpoints also evaluated for absolute extrema?
  3. How does the cubic root affect the function's behavior near zero?
  4. Can the second derivative test confirm the nature of the critical points here?
  5. How would you apply the same method to a different interval, e.g., [0,5][0, 5]?

Tip:

Always check both the critical points and the boundary points of the interval when looking for absolute extrema!

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Math Problem Analysis

Mathematical Concepts

Calculus
Critical Points
Derivatives
Extrema

Formulas

f(t) = t - ³√t
f'(t) = 1 - 1/(3√t²)

Theorems

First Derivative Test
Critical Point Theorem

Suitable Grade Level

College level (Calculus I)