Consider the function f: (0, ∞) arrow (-∞, ∞) given by f(x) = √x , _e(x) - x + 1 . Then which one of the following statements is TRUE?
Options
- AThe derivative of the function f is decreasing in the interval (0, 1)
- BThe function f has a local maximum at some point a ∈ (0, ∞)
- CThe function f has a local minimum at some point b ∈ (0, ∞)
- DThe function f has NEITHER a point of local maximum NOR a point of local minimum in the interval (0, ∞)
Correct answer
D. The function f has NEITHER a point of local maximum NOR a point of local minimum in the interval (0, ∞)
Step-by-step solution
Given f(x) = √x _e(x) - x + 1 Differentiating with respect to x : f'(x) = 1 2√x _e(x) + √x 1 x - 1 f'(x) = _e(x) 2√x + 1 √x - 1 Differentiating again with respect to x : f''(x) = d dx 1 2 x^ -1/2 _e(x) + x^ -1/2 - 1 f''(x) = 1 2 - 1 2 x^ -3/2 _e(x) + x^ -1/2 · 1 x - 1 2 x^ -3/2 f''(x) = - _e(x) 4x^ 3/2 + 1 2x^ 3/2 - 1 2x^ 3/2 f''(x) = - _e(x) 4x^ 3/2 For x ∈ (0, 1) , _e(x) 0 . Thus, f'(x) is strictly increasing in (0, 1) . For x ∈ (1, ∞) , _e(x) > 0 , which implies f''(x) Therefore, f'(x) attains its maximum value at x = 1 . Maximum value of f'(x) = f'(1) = _e(1) 2 + 1 - 1 = 0 . Since the maximum value of f'(x) is 0 , we have f'(x) ≤ 0 for all x ∈ (0, ∞) , with equality holding only at x = 1 . This means f(x) is a strictly decreasing function on (0, ∞) . Hence, f(x) has neither a point of local maximum nor a point of local minimum in the interval (0, ∞) . Answer: The function f has NEITHER a point of local maximum NOR a point of local minimum in the interval (0, ∞)