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**Illustration 1: Let f (x) = (4â€“x ^{2})^{2/3}, then f has a**

f'(x) = â€“4x / 3(4â€“x

The numerator gives the points of local maxima while the denominator gievs the points of local minima.

Hence, at x = â€“2 local minima

x = 0 local maxima

x = 2 local minima

**Illustration 2: The absolute minimum value of x ^{4} â€“ x^{2 }â€“ 2x+ 5**

Hence, f'(x) = 4x

= (x â€“ 1) (4x

Clearly at x = 1, we will set the minimum value which is 3.

Hence, (B) is the correct answer.

**Illustration 3: If f(x) = x ^{3 }+bx^{2} + cx+ d and 0 < b^{2}< c, then in (-âˆž, âˆž) (2004)**

fâ€™(x) = 3x

Now, D = 4b

Hence, D < 0

Therefore, fâ€™(x) = 3x

Hence, f(x) is a strictly increasing function.

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