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Application of Derivatives - Maxima and minima (first and second derivative test)

Grade 12CBSE

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

🔑Concepts

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Local Maxima and Minima: A function f(x)f(x) has a local maximum at x=cx=c if the value f(c)f(c) is the highest in its immediate neighborhood. Graphically, this appears as the peak of a 'hill' where the function stops increasing and starts decreasing. Conversely, a local minimum occurs at x=cx=c if f(c)f(c) is the lowest value in its neighborhood, appearing as the bottom of a 'valley' where the function stops decreasing and starts increasing.

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Critical Points: These are points x=cx=c in the domain of ff where either f′(c)=0f'(c) = 0 or the function is not differentiable. On a graph, these points represent potential turning points where the tangent line is horizontal (slope is zero) or where the graph has a sharp corner or 'kink'.

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First Derivative Test: This test uses the sign of f′(x)f'(x) to identify extrema. If f′(x)f'(x) changes from positive (++) to negative (−-) as xx increases through cc, then cc is a local maximum (climbing up then sliding down). If f′(x)f'(x) changes from negative (−-) to positive (++), then cc is a local minimum (sliding down then climbing up). If the sign of f′(x)f'(x) does not change, cc is neither a local maximum nor a local minimum.

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Second Derivative Test: This is often a quicker way to classify critical points. If f′(c)=0f'(c) = 0 and f′′(c)<0f''(c) < 0, the function is 'concave down' (shaped like an upside-down bowl) at cc, making it a local maximum. If f′(c)=0f'(c) = 0 and f′′(c)>0f''(c) > 0, the function is 'concave up' (shaped like a bowl) at cc, making it a local minimum. If f′′(c)=0f''(c) = 0, the test is inconclusive.

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Point of Inflection: A point on the graph where the concavity changes (e.g., from concave up to concave down). Visually, the curve 'flattens out' and switches its bend. At these points, f′′(x)f''(x) is typically 00 or undefined, but the first derivative test shows no change in direction (max/min).

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Absolute Maxima and Minima: On a closed interval [a,b][a, b], the absolute maximum is the highest overall value and the absolute minimum is the lowest overall value. These can occur either at critical points within the interval or at the boundaries (endpoints aa and bb). Visually, it is the highest and lowest point on the restricted segment of the graph.

📐Formulae

f′(c)=0f'(c) = 0 (Condition for stationary points)

f′(x)>0  ⟹  Function is increasingf'(x) > 0 \implies \text{Function is increasing}

f′(x)<0  ⟹  Function is decreasingf'(x) < 0 \implies \text{Function is decreasing}

f′(c)=0 and f′′(c)<0  ⟹  x=c is a local maximumf'(c) = 0 \text{ and } f''(c) < 0 \implies x=c \text{ is a local maximum}

f′(c)=0 and f′′(c)>0  ⟹  x=c is a local minimumf'(c) = 0 \text{ and } f''(c) > 0 \implies x=c \text{ is a local minimum}

Absolute Extrema=max⁡/min⁡{f(a),f(b),f(c1),f(c2),… }\text{Absolute Extrema} = \max/\min \{f(a), f(b), f(c_1), f(c_2), \dots \}

💡Examples

Problem 1:

Find the local maxima and minima for the function f(x)=2x3−6x2+6x+5f(x) = 2x^3 - 6x^2 + 6x + 5 using the first derivative test.

Solution:

  1. Find the first derivative: f′(x)=6x2−12x+6f'(x) = 6x^2 - 12x + 6.\n2. Factorize the derivative: f′(x)=6(x2−2x+1)=6(x−1)2f'(x) = 6(x^2 - 2x + 1) = 6(x-1)^2.\n3. Set f′(x)=0f'(x) = 0 to find critical points: 6(x−1)2=0  ⟹  x=16(x-1)^2 = 0 \implies x = 1.\n4. Analyze the sign of f′(x)f'(x) around x=1x=1:\n- For x<1x < 1, say x=0x=0: f′(0)=6(0−1)2=6>0f'(0) = 6(0-1)^2 = 6 > 0 (Positive).\n- For x>1x > 1, say x=2x=2: f′(2)=6(2−1)2=6>0f'(2) = 6(2-1)^2 = 6 > 0 (Positive).\n5. Since f′(x)f'(x) does not change sign as xx increases through 11, x=1x=1 is neither a local maximum nor a local minimum. It is a point of inflection.

Explanation:

In this problem, we find the critical point where the slope is zero. By checking the sign of the derivative on both sides of the point, we observe no sign change (it stays positive), which visually means the graph flattens momentarily but continues to rise, indicating an inflection point rather than a peak or valley.

Problem 2:

Find the absolute maximum and minimum values of f(x)=x3−3xf(x) = x^3 - 3x on the closed interval [0,2][0, 2].

Solution:

  1. Find the derivative: f′(x)=3x2−3f'(x) = 3x^2 - 3.\n2. Find critical points: 3(x2−1)=0  ⟹  x=1,−13(x^2 - 1) = 0 \implies x = 1, -1.\n3. Identify points within the interval [0,2][0, 2]: Only x=1x = 1 is in the interval.\n4. Evaluate f(x)f(x) at the critical point and the endpoints x=0x=0 and x=2x=2:\n- f(0)=(0)3−3(0)=0f(0) = (0)^3 - 3(0) = 0\n- f(1)=(1)3−3(1)=−2f(1) = (1)^3 - 3(1) = -2\n- f(2)=(2)3−3(2)=8−6=2f(2) = (2)^3 - 3(2) = 8 - 6 = 2\n5. Comparing the values: The absolute maximum value is 22 (at x=2x=2) and the absolute minimum value is −2-2 (at x=1x=1).

Explanation:

To find absolute extrema on a closed interval, we must check both the interior critical points and the boundary points of the interval. We evaluate the function at these candidates and simply select the largest and smallest results.