Question 36 of 82intermediate🔧 ApplyShort Answer4 marks

Prove that the function given by f(x) = x³ - 3x² + 3x - 100 is increasing in R.

Correct Answer

Given f(x) = x³ - 3x² + 3x - 100. Differentiating with respect to x, we get: f'(x) = 3x² - 6x + 3

= 3(x² - 2x + 1)

= 3(x - 1)²

Since (x - 1)² ≥\geq 0 for all x ∈ R, we have f'(x) ≥\geq 0 for all x ∈ R.

By Theorem 1, if f'(x) ≥\geq 0 for each x in an interval, then f is increasing in that interval.

Therefore, the function f is increasing in R.

Exercise: EXERCISE 6.2 | Q: 18 | (Page 13)
For More Understanding

Explanation

This question follows the same pattern as Example 8 in the textbook. The key technique is to find the derivative, simplify it into a perfect square form, and show that it is always non-negative. Since f'(x) = 3(x-1)² ≥\geq 0 for all real x, by Theorem 1 (the first derivative test), the function is increasing throughout R. The constant term -100 does not affect the derivative and hence does not affect the increasing/decreasing nature of the function.

Solution Steps

  1. Step 1: Find the derivative f'(x) = 3x² - 6x + 3

  2. Step 2: Simplify: f'(x) = 3(x² - 2x + 1) = 3(x - 1)²

  3. Step 3: Note that (x - 1)² ≥\geq 0 for all x ∈ R, so f'(x) ≥\geq 0 for all x ∈ R

  4. Step 4: Apply Theorem 1 to conclude f is increasing in R

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