Solved Examples on Maxima & Minima

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

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

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.

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