The pattern and behavior of the data in any time series are based on which of the following components: (A) Secular trend component (B) Seasonal…

A function is strictly decreasing on an interval when the derivative is negative.

Given: $f(x) = x^4 - \frac{x^3}{3}$f(x)=x4−3x3​

Finding the derivative using the power rule:

$$f'(x) = 4x^3 - \frac{3x^2}{3}$$f′(x)=4x3−33x2​

$$f'(x) = 4x^3 - x^2$$f′(x)=4x3−x2

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Factoring out common terms:

$$f'(x) = x^2(4x - 1)$$f′(x)=x2(4x−1)

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Setting $f'(x) = 0$f′(x)=0:

$$x^2(4x - 1) = 0$$x2(4x−1)=0

This gives:

$x^2 = 0$x2=0 → $x = 0$x=0

$4x - 1 = 0$4x−1=0 → $x = \frac{1}{4}$x=41​

Critical points: $x = 0$x=0 and $x = \frac{1}{4}$x=41​

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Analyzing the sign of $f'(x) = x^2(4x - 1)$f′(x)=x2(4x−1) in each interval:

Note that $x^2$x2 is always non-negative. The sign of $f'(x)$f′(x) depends on $(4x - 1)$(4x−1).

For $x < \frac{1}{4}$x<41​:

$(4x - 1) < 0$(4x−1)<0, so $f'(x) \leq 0$f′(x)≤0

The function is decreasing.

For $x > \frac{1}{4}$x>41​:

$(4x - 1) > 0$(4x−1)>0, so $f'(x) > 0$f′(x)>0

The function is increasing.

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The function is strictly decreasing on $\left(-\infty, \frac{1}{4}\right)$(−∞,41​).