Math Problem Statement

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Find the extreme values of the function f on the interval [1.5, 4] for f(x) = x^8 + 8/x.

Solution

To find the absolute minimum and maximum values of the function f(x)=x8+8xf(x) = x^8 + \frac{8}{x} on the interval [1.5,4][1.5, 4], we need to follow these steps:

Step 1: Compute the Derivative

First, let's find the derivative f(x)f'(x) to identify the critical points within the interval.

f(x)=x8+8xf(x) = x^8 + \frac{8}{x}

The derivative f(x)f'(x) is:

f(x)=8x78x2f'(x) = 8x^7 - \frac{8}{x^2}

Step 2: Set the Derivative to Zero

We set f(x)=0f'(x) = 0 to find critical points.

8x78x2=08x^7 - \frac{8}{x^2} = 0

This can be rearranged to:

8x7=8x28x^7 = \frac{8}{x^2}

Dividing both sides by 8:

x7=1x2x^7 = \frac{1}{x^2}

Multiply both sides by x2x^2:

x9=1x^9 = 1

Taking the ninth root of both sides gives:

x=1x = 1

However, x=1x = 1 is not within the interval [1.5,4][1.5, 4], so we discard this solution.

Step 3: Evaluate f(x)f(x) at the Endpoints

Since there are no critical points in the interval, we evaluate the function at the endpoints x=1.5x = 1.5 and x=4x = 4.

  1. At x=1.5x = 1.5: f(1.5)=(1.5)8+81.5f(1.5) = (1.5)^8 + \frac{8}{1.5} Calculating each part:

    • (1.5)8=25.62890625(1.5)^8 = 25.62890625
    • 81.55.3333\frac{8}{1.5} \approx 5.3333

    So, f(1.5)25.62890625+5.3333=30.9622f(1.5) \approx 25.62890625 + 5.3333 = 30.9622

  2. At x=4x = 4: f(4)=(4)8+84f(4) = (4)^8 + \frac{8}{4} Calculating each part:

    • (4)8=65536(4)^8 = 65536
    • 84=2\frac{8}{4} = 2

    So, f(4)=65536+2=65538f(4) = 65536 + 2 = 65538

Conclusion

  • The absolute minimum value of f(x)f(x) on the interval [1.5,4][1.5, 4] is 30.962230.9622 at x=1.5x = 1.5.
  • The absolute maximum value of f(x)f(x) on the interval [1.5,4][1.5, 4] is 6553865538 at x=4x = 4.

Would you like more details on this process or any specific step?

Follow-up Questions

  1. What is the importance of checking endpoints when finding absolute extrema on a closed interval?
  2. How would the process differ if the interval were open (e.g., (1.5,4)(1.5, 4))?
  3. What would happen if a critical point did exist within the interval?
  4. How do derivatives help identify local maxima and minima in general?
  5. Can you explain why x=1x = 1 was not considered a valid solution in this interval?

Tip: When solving for extrema on a closed interval, always check both critical points and endpoints.

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Math Problem Analysis

Mathematical Concepts

Calculus
Optimization
Critical Points
Absolute Maximum and Minimum

Formulas

f(x) = x^8 + 8/x
f'(x) = 8x^7 - 8/x^2

Theorems

Extreme Value Theorem

Suitable Grade Level

College Calculus