Question 47 of 82intermediate🔧 ApplyNumerical4 marks

Find the maximum value of 2x³ - 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1].

Correct Answer

(interval_-3_-1) Maximum value in [-3, -1] is 139

(interval_1_3) Maximum value in [1, 3] is 89

Exercise: EXERCISE 6.3 | Q: 10 | (Page 29)
For More Understanding

Explanation

Following the method from Example 27 in the textbook, we find critical points by setting f'(x) = 0, then evaluate the function at critical points within the interval and at the endpoints. The absolute maximum is the largest of these values.

Solution Steps

  1. Step 1: Given f(x) = 2x³ - 24x + 107. Find f'(x) = 6x² - 24 = 6(x - 2)(x + 2)

  2. Step 2: Critical points: f'(x) = 0 gives x = 2 and x = -2

  3. Step 3: For interval [1, 3]: Critical point x = 2 lies in interval. Evaluate f(1) = 2(1)³ - 24(1) + 107 = 85; f(2) = 2(2)³ - 24(2) + 107 = 75; f(3) = 2(3)³ - 24(3) + 107 = 89

  4. Step 4: Maximum value in [1, 3] is 89 at x = 3

  5. Step 5: For interval [-3, -1]: Critical point x = -2 lies in interval. Evaluate f(-3) = 2(-3)³ - 24(-3) + 107 = 125; f(-2) = 2(-2)³ - 24(-2) + 107 = 139; f(-1) = 2(-1)³ - 24(-1) + 107 = 129

  6. Step 6: Maximum value in [-3, -1] is 139 at x = -2

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