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Calculus - Monotony – max, min

Grade 11IB_AA

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

🔑Concepts

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A function f(x)f(x) is strictly increasing on an interval if f′(x)>0f'(x) > 0 for all xx in that interval.

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A function f(x)f(x) is strictly decreasing on an interval if f′(x)<0f'(x) < 0 for all xx in that interval.

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Stationary points occur where the gradient is zero, i.e., f′(x)=0f'(x) = 0. These points can be local maxima, local minima, or stationary points of inflection.

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The First Derivative Test: If f′(x)f'(x) changes sign from positive to negative at x=cx = c, then f(c)f(c) is a local maximum. If f′(x)f'(x) changes sign from negative to positive at x=cx = c, then f(c)f(c) is a local minimum.

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The Second Derivative Test: If f′(c)=0f'(c) = 0 and f′′(c)<0f''(c) < 0, the point is a local maximum. If f′(c)=0f'(c) = 0 and f′′(c)>0f''(c) > 0, the point is a local minimum.

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Global (Absolute) Extrema: To find the global maximum or minimum on a closed interval [a,b][a, b], compare the values of f(x)f(x) at all stationary points within the interval and at the endpoints f(a)f(a) and f(b)f(b).

📐Formulae

f′(x)=0 (Condition for stationary points)f'(x) = 0 \text{ (Condition for stationary points)}

f′(x)>0  ⟹  Increasingf'(x) > 0 \implies \text{Increasing}

f′(x)<0  ⟹  Decreasingf'(x) < 0 \implies \text{Decreasing}

f′′(c)<0 at f′(c)=0  ⟹  Local Maximumf''(c) < 0 \text{ at } f'(c)=0 \implies \text{Local Maximum}

f′′(c)>0 at f′(c)=0  ⟹  Local Minimumf''(c) > 0 \text{ at } f'(c)=0 \implies \text{Local Minimum}

💡Examples

Problem 1:

Find the intervals of increase and decrease for the function f(x)=x3−3x2−9x+5f(x) = x^3 - 3x^2 - 9x + 5.

Solution:

  1. Find the first derivative: f′(x)=3x2−6x−9f'(x) = 3x^2 - 6x - 9.
  2. Set f′(x)=0f'(x) = 0: 3(x2−2x−3)=0  ⟹  3(x−3)(x+1)=03(x^2 - 2x - 3) = 0 \implies 3(x-3)(x+1) = 0.
  3. Critical values are x=3x = 3 and x=−1x = -1.
  4. Test intervals: For x<−1x < -1, f′(−2)=3(−5)(−1)=15>0f'(-2) = 3(-5)(-1) = 15 > 0 (Increasing). For −1<x<3-1 < x < 3, f′(0)=−9<0f'(0) = -9 < 0 (Decreasing). For x>3x > 3, f′(4)=3(1)(5)=15>0f'(4) = 3(1)(5) = 15 > 0 (Increasing). Intervals: Increasing on (−∞,−1)∪(3,∞)(-\infty, -1) \cup (3, \infty); Decreasing on (−1,3)(-1, 3).

Explanation:

The sign of the derivative f′(x)f'(x) determines whether the function is going up or down. We find the roots of the derivative to identify where the direction might change.

Problem 2:

Determine the coordinates and the nature of the stationary points for f(x)=x+4xf(x) = x + \frac{4}{x} for x≠0x \neq 0.

Solution:

  1. Differentiate: f(x)=x+4x−1  ⟹  f′(x)=1−4x−2=1−4x2f(x) = x + 4x^{-1} \implies f'(x) = 1 - 4x^{-2} = 1 - \frac{4}{x^2}.
  2. Set f′(x)=0f'(x) = 0: 1=4x2  ⟹  x2=4  ⟹  x=2,−21 = \frac{4}{x^2} \implies x^2 = 4 \implies x = 2, -2.
  3. Find yy-coordinates: f(2)=2+2=4f(2) = 2 + 2 = 4; f(−2)=−2−2=−4f(-2) = -2 - 2 = -4.
  4. Second derivative: f′′(x)=8x−3=8x3f''(x) = 8x^{-3} = \frac{8}{x^3}.
  5. Test x=2x=2: f′′(2)=88=1>0  ⟹  f''(2) = \frac{8}{8} = 1 > 0 \implies Local Minimum at (2,4)(2, 4).
  6. Test x=−2x=-2: f′′(−2)=8−8=−1<0  ⟹  f''(-2) = \frac{8}{-8} = -1 < 0 \implies Local Maximum at (−2,−4)(-2, -4).

Explanation:

The Second Derivative Test is used here. A positive second derivative indicates the graph is concave up (a valley/minimum), while a negative second derivative indicates it is concave down (a hill/maximum).