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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 called strictly increasing on an interval if for any two points x1,x2x_1, x_2 in that interval, x1<x2x_1 < x_2 implies f(x1)<f(x2)f(x_1) < f(x_2). Geometrically, the slope of the tangent at any point, given by f′(x)f'(x), is positive (f′(x)>0f'(x) > 0).

Graph of a strictly increasing function where the curve rises from left to right.
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A function f(x)f(x) is called strictly decreasing on an interval if x1<x2x_1 < x_2 implies f(x1)>f(x2)f(x_1) > f(x_2). In this case, the derivative f′(x)f'(x) is negative (f′(x)<0f'(x) < 0), meaning the tangent to the curve makes an obtuse angle with the positive x-axis.

Graph of a strictly decreasing function where the curve falls from left to right.
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Critical points occur where f′(x)=0f'(x) = 0 or where f(x)f(x) is not differentiable. These points help in dividing the domain into intervals where the function maintains a constant sign for its derivative.

Graph showing a local minimum where the tangent is horizontal, indicating a critical point.
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The First Derivative Test: If f′(x)>0f'(x) > 0 for all x∈(a,b)x \in (a, b), then ff is strictly increasing in [a,b][a, b]. If f′(x)<0f'(x) < 0 for all x∈(a,b)x \in (a, b), then ff is strictly decreasing in [a,b][a, b].

📐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.

Problem 3:

Find the intervals in which the function f(x)=x2−4x+6f(x) = x^2 - 4x + 6 is (a) strictly increasing, (b) strictly decreasing.

Parabola opening upwards with vertex at x=2, showing decreasing behavior before x=2 and increasing after.

Solution:

  1. Find the derivative: f′(x)=2x−4f'(x) = 2x - 4.
  2. Set f′(x)=0f'(x) = 0 to find critical points: 2x−4=0  ⟹  x=22x - 4 = 0 \implies x = 2.
  3. The point x=2x = 2 divides the real line into two intervals: (−∞,2)(-\infty, 2) and (2,∞)(2, \infty).
  4. For x∈(−∞,2)x \in (-\infty, 2), let x=0x = 0: f′(0)=2(0)−4=−4<0f'(0) = 2(0) - 4 = -4 < 0. So, ff is strictly decreasing in (−∞,2)(-\infty, 2).
  5. For x∈(2,∞)x \in (2, \infty), let x=3x = 3: f′(3)=2(3)−4=2>0f'(3) = 2(3) - 4 = 2 > 0. So, ff is strictly increasing in (2,∞)(2, \infty).

Explanation:

We use the derivative to determine the slope. The critical point x=2x=2 is where the function changes from decreasing to increasing.

Problem 4:

Determine the intervals in which f(x)=4x3−6x2−72x+30f(x) = 4x^3 - 6x^2 - 72x + 30 is strictly increasing or decreasing.

Cubic function graph showing peaks and valleys at x=-2 and x=3 respectively.

Solution:

  1. f′(x)=12x2−12x−72=12(x2−x−6)f'(x) = 12x^2 - 12x - 72 = 12(x^2 - x - 6).
  2. Factorizing: f′(x)=12(x−3)(x+2)f'(x) = 12(x - 3)(x + 2).
  3. Critical points: x=3,−2x = 3, -2.
  4. Intervals: (−∞,−2)(-\infty, -2), (−2,3)(-2, 3), and (3,∞)(3, \infty).
  5. In (−∞,−2)(-\infty, -2), f′(x)>0f'(x) > 0 (Strictly Increasing).
  6. In (−2,3)(-2, 3), f′(x)<0f'(x) < 0 (Strictly Decreasing).
  7. In (3,∞)(3, \infty), f′(x)>0f'(x) > 0 (Strictly Increasing).

Explanation:

By checking the signs of the factors (x−3)(x-3) and (x+2)(x+2) in each interval, we determine the overall sign of the derivative.