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प्रश्न
Determine for which values of x, the function y = `x^4 – (4x^3)/3` is increasing and for which values, it is decreasing.
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उत्तर
y = `x^4 – (4x^3)/3`
⇒ `"dy"/"dx"` = 4x3 – 4x2
= 4x2(x – 1)
Now, `"dy"/"dx"` = 0
⇒ x = 0, x = 1.
Since f′(x) < 0 ∀ ∈x `(- oo, 0)` ∪ (0, 1) and f is continuous in `(- oo, 0]` and [0, 1].
Therefore f is decreasing in `(- oo, 1]` and f is increasing in `[1, oo)`.
Note: Here f is strictly decreasing in `(- oo, 0)` ∪ (0, 1) and is strictly increasing in `(1, oo)`.
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