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Application of Derivatives - Increasing and decreasing functions

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A function f(x)f(x) is said to be strictly increasing on an open interval (a,b)(a, b) if for any two numbers x1x_1 and x2x_2 in the interval, x1<x2x_1 < x_2 implies f(x1)<f(x2)f(x_1) < f(x_2). Visually, the graph of the function rises continuously from left to right without any flat regions.

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A function f(x)f(x) is strictly decreasing on an open interval (a,b)(a, b) if for any two numbers x1x_1 and x2x_2 in the interval, x1<x2x_1 < x_2 implies f(x1)>f(x2)f(x_1) > f(x_2). Geometrically, the curve slopes downward continuously as the x-values increase.

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The derivative f′(x)f'(x) determines the slope of the tangent. If f′(x)>0f'(x) > 0 for all xx in (a,b)(a, b), the function is strictly increasing. On a graph, this corresponds to every tangent line having a positive angle with the positive x-axis.

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If f′(x)<0f'(x) < 0 for all xx in (a,b)(a, b), the function is strictly decreasing. This indicates that the slope of the tangent at every point in the interval is negative, reflecting a downward trend in the graph.

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Critical points are values of xx where f′(x)=0f'(x) = 0 or where the function is not differentiable. On a graph, these are often the 'turning points' where a curve transitions from climbing to falling (a peak) or falling to climbing (a valley).

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To determine the intervals of monotonicity, we find the critical points and divide the domain into sub-intervals. We then test the sign of f′(x)f'(x) in each sub-interval. A change in the sign of f′(x)f'(x) around a critical point identifies where the function changes direction.

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A function is non-decreasing (increasing) if f′(x)≥0f'(x) \ge 0 and non-increasing (decreasing) if f′(x)≤0f'(x) \le 0. The inclusion of the equality 'equal to zero' allows for horizontal segments in the graph where the function value stays constant for a stretch.

📐Formulae

Strictly Increasing: f′(x)>0f'(x) > 0 for x∈(a,b)x \in (a, b)

Strictly Decreasing: f′(x)<0f'(x) < 0 for x∈(a,b)x \in (a, b)

Increasing: f′(x)≥0f'(x) \ge 0 for x∈(a,b)x \in (a, b)

Decreasing: f′(x)≤0f'(x) \le 0 for x∈(a,b)x \in (a, b)

Constant Function: f′(x)=0f'(x) = 0 for all x∈(a,b)x \in (a, b)

💡Examples

Problem 1:

Find the intervals in which the function f(x)=2x2−3xf(x) = 2x^2 - 3x is strictly increasing or strictly decreasing.

Solution:

  1. Find the derivative: f′(x)=ddx(2x2−3x)=4x−3f'(x) = \frac{d}{dx}(2x^2 - 3x) = 4x - 3.
  2. Set f′(x)=0f'(x) = 0 to find the critical point: 4x−3=0⇒x=344x - 3 = 0 \Rightarrow x = \frac{3}{4}.
  3. The point x=34x = \frac{3}{4} divides the real line into two intervals: (−∞,34)(-\infty, \frac{3}{4}) and (34,∞)(\frac{3}{4}, \infty).
  4. Test interval (−∞,34)(-\infty, \frac{3}{4}): Let x=0x = 0. f′(0)=4(0)−3=−3<0f'(0) = 4(0) - 3 = -3 < 0. Thus, ff is strictly decreasing on (−∞,34)(-\infty, \frac{3}{4}).
  5. Test interval (34,∞)(\frac{3}{4}, \infty): Let x=1x = 1. f′(1)=4(1)−3=1>0f'(1) = 4(1) - 3 = 1 > 0. Thus, ff is strictly increasing on (34,∞)(\frac{3}{4}, \infty).

Explanation:

We use the First Derivative Test. By finding where the slope is positive or negative, we identify the direction of the function.

Problem 2:

Prove that the function f(x)=cos⁡xf(x) = \cos x is strictly decreasing in (0,π)(0, \pi).

Solution:

  1. Differentiate the function: f′(x)=ddx(cos⁡x)=−sin⁡xf'(x) = \frac{d}{dx}(\cos x) = -\sin x.
  2. Analyze the sign of f′(x)f'(x) in the given interval (0,π)(0, \pi).
  3. In the first and second quadrants (i.e., 0<x<π0 < x < \pi), the value of sin⁡x\sin x is always positive: sin⁡x>0\sin x > 0.
  4. Therefore, f′(x)=−sin⁡xf'(x) = -\sin x will be negative: f′(x)<0f'(x) < 0 for all x∈(0,π)x \in (0, \pi).
  5. Since f′(x)<0f'(x) < 0, the function is strictly decreasing on (0,π)(0, \pi).

Explanation:

Trigonometric functions are evaluated by checking the sign of their derivatives within specific quadrants of the unit circle.