Illustration 1: Let f (x) = (4–x2)2/3, then f has a
(A) a local maxima at x = 0
(B) a local maxima at x = 2
(C) a local maxima at x = –2
(D) none of these
Solution: The given function is f(x) = (4 – x2)2/3
f'(x) = –4x / 3(4–x2)1/3 = 4x / 3(x2–4)1/3
The numerator gives the points of local maxima while the denominator gievs the points of local minima.
Hence, at x = –2 local minima
x = 0 local maxima
x = 2 local minima
Illustration 2: The absolute minimum value of x4 – x2 – 2x+ 5
(A) is equal to 5
(B) is equal to 3
(C) is equal to 7
(D) does not exist
Solution: The given function is x4 – x2 – 2x+ 5.
Hence, f'(x) = 4x3 – 2x – 2
= (x – 1) (4x2 + 4x + 2)
Clearly at x = 1, we will set the minimum value which is 3.
Hence, (B) is the correct answer.
Illustration 3: If f(x) = x3 +bx2 + cx+ d and 0 < b2< c, then in (-∞, ∞) (2004)
1. f(x) is strictly increasing function
2. f(x) has local maxima
3. f(x) is strictly decreasing function
4. f(x) is bounded
Solution: Givenf(x) = x3 +bx2 + cx+ d
f’(x) = 3x2 +2bx + c
Now, D = 4b2 -12c
Hence, D < 0
Therefore, f’(x) = 3x2 +2bx + c > 0 for all x in (-∞, ∞)
Hence, f(x) is a strictly increasing function.