Question 29 of 82intermediate🔧 ApplyShort Answer4 marks

Prove that the function f given by f(x) = x² - x + 1 is neither strictly increasing nor decreasing on (-1, 1).

Correct Answer

Given f(x) = x² - x + 1.

Step 1: Find the derivative. f'(x) = 2x - 1

Step 2: Find critical point. f'(x) = 0 gives 2x - 1 = 0, so x = 1/2.

Step 3: Analyze sign of f'(x) on intervals. The point x = 1/2 divides the real line into two intervals: (-∞, 1/2) and (1/2, ∞).

In (-∞, 1/2): f'(x) = 2x - 1 < 0, so f is decreasing.

In (1/2, ∞): f'(x) = 2x - 1 > 0, so f is increasing.

Step 4: Analyze on (-1, 1). The interval (-1, 1) contains x = 1/2.

  • On (-1, 1/2): f'(x) < 0, so f is decreasing.
  • On (1/2, 1): f'(x) > 0, so f is increasing.

Since f is decreasing on one part and increasing on another part of (-1, 1), the function f is neither strictly increasing nor strictly decreasing on (-1, 1).

Exercise: EXERCISE 6.2 | Q: 11 | (Page 13)
For More Understanding

Explanation

This question follows the same pattern as Example 9(c) and Example 10 in the textbook. The key insight is that when a critical point (where f'(x)=0) lies within an interval, the function changes behavior across that point. Since x=1/2 lies inside (-1,1), the function decreases before this point and increases after it, making it neither strictly increasing nor strictly decreasing on the entire interval.

Solution Steps

  1. Step 1: Find f'(x) = 2x - 1

  2. Step 2: Solve f'(x) = 0 to get x = 1/2

  3. Step 3: Determine sign of f'(x) on intervals (-∞, 1/2) and (1/2, ∞)

  4. Step 4: Observe that (-1, 1) contains x = 1/2, so f decreases on (-1, 1/2) and increases on (1/2, 1)

  5. Step 5: Conclude f is neither strictly increasing nor strictly decreasing on (-1, 1)

PreviousNext