Application of Derivatives · Question 2

Let R denote the set of all real numbers. Define the function f: R R by f(x) = cases 2 - 2x2 - x2 1x & if x 0, \ 2 & if x = 0. cases Then which one of the following statements is TRUE?

1. The function f is NOT differentiable at x = 0
2. There is a positive real number , such that f is a decreasing function on the interval (0, )
3. For any positive real number , the function f is NOT an increasing function on the interval (- , 0)
4. x = 0 is a point of local minima of f

Answer

(3) For any positive real number , the function f is NOT an increasing function on the interval (- , 0)

Solution

f'(0) = x 0 -2x2 - x2 (1/x)x = x 0 (-2x - x 1x ) = 0, so f is differentiable at 0; (A) is false.

For x 0: f'(x) = -4x - 2x 1x + 1x.

As x 0, -4x - 2x 1x 0 while 1x keeps taking the values 1 and -1 in every interval (0, ) and (- ,0).

So f' takes both signs in every such interval: f is neither increasing nor decreasing there. (B) false, (C) true.

f(x) - f(0) = -x2 (2 + 1x ) < 0 for x 0, so x = 0 is a local maximum; (D) false.

Answer: (C)