Calculus: Maxima, Minima, and Concavity Analysis

Definitions of Increasing, Decreasing, and Constant Functions

• Let $f$ be defined on an interval and let $x_1$ and $x_2$ denote points in that interval.

• Strictly Increasing: $f$ is strictly increasing on the interval if f(x_1) < f(x_2) whenever x_1 < x_2.

• Strictly Decreasing: $f$ is strictly decreasing on the interval if f(x_1) > f(x_2) whenever x_1 < x_2.

• Constant: $f$ is constant on the interval if $f(x_1) = f(x_2)$ for all points $x_1$ and $x_2$.

Theorem for Monotonicity

Let $f$ be a function that is continuous in a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$.

• (a) If f'(x) > 0 for every value of $x$ in $(a, b)$, then $f$ is increasing on $[a, b]$.

• (b) If f'(x) < 0 for every value of $x$ in $(a, b)$, then $f$ is decreasing on $[a, b]$.

• (c) If $f'(x) = 0$ for every value of $x$ in $(a, b)$, then $f$ is constant on $[a, b]$.

Summary of Derivative Signs

• Increasing: f'(x) > 0

• Decreasing: f'(x) < 0

• Constant: $f'(x) = 0$

• Strictly increasing: f'(x) > 0

• Strictly decreasing: f'(x) < 0

• Increasing (Non-strict): $f'(x) \geq 0$

• Decreasing (Non-strict): $f'(x) \leq 0$

• The Case of $\sin(x)$ on $[0, \pi]$: For $y = \sin(x)$, it is neither strictly increasing nor decreasing over the full interval $[0, \pi]$, as it increases on $[0, \frac{\pi}{2}]$ and decreases on $[\frac{\pi}{2}, \pi]$.

Problems on Finding Intervals of Monotonicity

Problem (a): $f(x) = x^2 - 4x + 3$

1. Given: $f(x) = (x - 2)^2 - 1$.

2. Derivative: $f'(x) = 2x - 4$.

3. For increasing function: f'(x) > 0 \Rightarrow 2x - 4 > 0 \Rightarrow x > 2. Interval: $[2, \infty)$.

4. For decreasing function: f'(x) < 0 \Rightarrow 2(x - 2) < 0 \Rightarrow x < 2. Interval: $(-\infty, 2]$.

Problem (b): $f(x) = x^3 - 3x^2 + 4$

1. Derivative: $f'(x) = 3x^2 - 6x$.

2. Critical values: Setting $3x^2 - 6x = 0 \Rightarrow 3x(x - 2) = 0$, giving $x = 0, 2$.

3. Sign analysis:

   • If x < 0, f'(x) > 0 (Increasing).

   • If 0 < x < 2, f'(x) < 0 (Decreasing).

   • If x > 2, f'(x) > 0 (Increasing).

4. Conclusion: $f$ is increasing on $(-\infty, 0] \cup [2, \infty)$ and decreasing on $[0, 2]$.

Problem (c): $f(x) = x^3 - 12x + 1$

1. Derivative: $f'(x) = 3x^2 - 12 = 3(x^2 - 4)$.

2. Critical values: $x = 2, -2$.

3. Intervals Analysis:

   • Interval x < -2: Term $(x - 2)$ is negative, $(x + 2)$ is negative, so f'(x) > 0 (Increasing).

   • Interval -2 < x < 0: Negative sign product, f'(x) < 0 (Decreasing).

   • Interval 0 < x < 2: Negative sign product, f'(x) < 0 (Decreasing).

   • Interval x > 2: Both positive, f'(x) > 0 (Increasing).

4. Conclusion: $f$ is increasing on $(-\infty, -2] \cup [2, \infty)$ and decreasing on $[-2, 0] \cup [0, 2]$.

Problem (d): $f(x) = x^3$

1. Derivative: $f'(x) = 3x^2$.

2. Analysis: Since 3x^2 > 0 for all $x \neq 0$, and at $x=0$, $f'(0)=0$.

3. Conclusion: $f$ is increasing on $(-\infty, 0] \cup [0, \infty)$.

Concavity Definitions and Theorems

• Concavity Definition: Let $f$ be differentiable on an interval.

  • (a) $f$ is called concave up on the interval if $f'(x)$ is increasing on the interval.

  • (b) $f$ is called concave down on the interval if $f'(x)$ is decreasing on the interval.

• Second Derivative Theorem:

  • (a) If f''(x) > 0 on an open interval, then $f$ is concave up on $(a, b)$.

  • (b) If f''(x) < 0 on an open interval, then $f$ is concave down on $(a, b)$.

Problems on Finding Intervals of Concavity

Problem 1: $f(x) = x^3$

1. First derivative: $f'(x) = 3x^2$.

2. Second derivative: $f''(x) = 6x$.

3. Concave up: 6x > 0 \Rightarrow x > 0. Interval: $(0, \infty)$.

4. Concave down: 6x < 0 \Rightarrow x < 0. Interval: $(-\infty, 0)$.

Problem 2: $f(x) = 3x^3 - 4x^2 + 6$

1. First derivative: $f'(x) = 9x^2 - 8x$.

2. Second derivative: $f''(x) = 18x - 8 = 2(9x - 4)$.

3. Concave up: 2(9x - 4) > 0 \Rightarrow x > \frac{4}{9}. Interval: $(\frac{4}{9}, \infty)$.

4. Concave down: 2(9x - 4) < 0 \Rightarrow x < \frac{4}{9}. Interval: $(-\infty, \frac{4}{9})$.

Problem 3: $f(x) = x^3 - 12x + 4$

1. Second derivative result ($f''(x) = 6x$).

2. Concave up: $(0, \infty)$.

3. Concave down: $(-\infty, 0)$.

Nature of Stationary and Critical Points

• Critical Value: The values of $x$ for which $\frac{dy}{dx} = 0$ or $f'(x) = 0$.

• Critical Point Classification:

  • Stationary point: Where $f'(x) = 0$.

  • Non-differentiable point: Where $f'(x)$ does not exist.

Theorem (First Derivative Test)

Suppose $f$ is continuous at a critical point $x_0$.

• (a) If f'(x) > 0 to the left of $x_0$ and f'(x) < 0 to the right of $x_0$, then $f$ has a relative maximum at $x_0$.

• (b) If f'(x) < 0 to the left of $x_0$ and f'(x) > 0 to the right of $x_0$, then $f$ has a relative minimum at $x_0$.

• (c) If $f'(x)$ has the same sign on both sides of $x_0$, then $f$ does not have a relative extremum (this is an inflection point).

• Note: Relative maximum or minimum are collectively known as extremum.

Practice Problems: First Derivative Test

Problem 1: $f(x) = x^2 - 2x + 5$

1. Derivative: $f'(x) = 2x - 2$.

2. Stationary point: $2x - 2 = 0 \Rightarrow x = 1$.

3. Analysis near $x = 1$:

   • at x < 1, $f'(x) = -ve$.

   • at x > 1, $f'(x) = +ve$.

4. Result: Minimum at $x = 1$.

Problem 2: $y = (x - 1)^4(x + 2)^3$

1. Derivative: $\frac{dy}{dx} = (x - 1)^4 \cdot 3(x + 2)^2 \cdot 1 + 4(x - 1)^3 \cdot (x + 2)^3 \cdot 1 = (x - 1)^3(x + 2)^2(7x + 5)$.

2. Stationary points: $x = 1, -2, -\frac{5}{7}$.

3. Nature of $x = 1$:

   • x < 1 \Rightarrow (dy/dx) = (-)(+)(+) = -

   • x > 1 \Rightarrow (dy/dx) = (+)(+)(+) = +

   • Result: Minimum at $x = 1$.

4. Nature of $x = -2$:

   • x < -2 \Rightarrow (dy/dx) = (-)(+)(-) = +

   • x > -2 \Rightarrow (dy/dx) = (-)(+)(-) = +

   • Result: Point of inflection at $x = -2$.

5. Nature of $x = -\frac{5}{7}$:

   • x < -\frac{5}{7} \Rightarrow (dy/dx) = (-)(+)(-) = +

   • x > -\frac{5}{7} \Rightarrow (dy/dx) = (-)(+)(+) = -

   • Result: Maximum at $x = -\frac{5}{7}$.

Homework Exercises

1. If $y = (x - 3)^4(x - 2)^5$. Answers: $x = 2$ (inflection), $x = 3$ (min), $x = -\frac{2}{5}$ (maximum).

2. If $y = (2x + 3)^3(3x - 1)^2$. Answers: $x = -\frac{3}{2}$ (inflection), $x = \frac{1}{3}$ (min), $x = -\frac{2}{5}$ (maximum).

Theorem: Second Derivative Test

Suppose $f$ is twice differentiable at stationary point $x_0$.

• (a) If f''(x_0) > 0, then $f$ has a relative minimum at $x_0$.

• (b) If f''(x_0) < 0, then $f$ has a relative maximum at $x_0$.

Extended Derivative Test Theorem

If $f''(x) = f'''(x) = f^{iv}(x) = \dots = f^{n-1} = 0$ and $f^n(x) \neq 0$:

• If $n$ is even, then $f(x)$ has either a maximum or a minimum.

• If $n$ is odd, no maximum or minimum value exists at that point.

Finding Max/Min Values using Second Derivative

Problem 1: $f(x) = x^3 - 12x + 4$

1. Derivative: $f'(x) = 3x^2 - 12$.

2. Tripical (Critical) points: $3(x^2 - 4) = 0 \Rightarrow x = \pm 2$.

3. Second derivative: $\frac{d^2y}{dx^2} = 6x$.

4. At $x = 2$: \frac{d^2y}{dx^2} = 12 > 0. Minimum at $x = 2$. $y_{min} = 2^3 - 12(2) + 4 = -12$.

5. At $x = -2$: \frac{d^2y}{dx^2} = -12 < 0. Maximum at $x = -2$. $y_{max} = (-2)^3 - 12(-2) + 4 = 20$.

Problem 2: $f(x) = 4 \sin(x) \cos(2x)$

1. Derivative: $f'(x) = 4 \cos^3(x) - 8 \sin^2(x) \cos(x)$.

2. Condition $f'(x) = 0 \Rightarrow 4 \cos^3(x)(1 - 2 \tan^2(x)) = 0$.

3. Roots: $\cos(x) = 0 \Rightarrow x = \frac{\pi}{2}$; $\tan(x) = \frac{1}{\sqrt{2}} \Rightarrow x = \tan^{-1}(\frac{1}{\sqrt{2}})$.

4. Second derivative: $\frac{d^2y}{dx^2} = 8 \sin^3(x) - 28 \cos^2(x) \sin(x)$.

5. At $x = \frac{\pi}{2}$: \frac{d^2y}{dx^2} = 8 > 0. Minimum value $y_{min} = 0$.

6. At $x = \tan^{-1}(\frac{1}{\sqrt{2}})$: $\cos(x) = \frac{\sqrt{2}}{\sqrt{3}}$, $\sin(x) = \frac{1}{\sqrt{3}}$. \frac{d^2y}{dx^2} = -\frac{48}{\sqrt{3}} < 0. Maximum value $y_{max} = 4 \cdot (\frac{1}{\sqrt{3}}) (\frac{\sqrt{2}}{\sqrt{3}})^2 = \frac{8}{3\sqrt{3}}$.

Problem 3: $f(x) = x^5 - 5x^4 + 5x^3 - 1$

1. $f'(x) = 5x^4 - 20x^3 + 15x^2 = 0 \Rightarrow x = 0, 1, 3$.

2. $f''(x) = 20x^3 - 60x^2 + 30x$; $f'''(x) = 60x^2 - 120x + 30$.

3. At $x = 0$: $f''(0) = 0, f'''(0) = 30 \neq 0$. No max/min.

4. At $x = 1$: f''(1) = -10 < 0. Max occurs; $y_{max} = 0$.

5. At $x = 3$: f''(30) = 90 > 0. Min occurs; $y_{min} = -28$.

Problem 4: $f(x) = 5x^6 - 18x^5 + 15x^4 - 10$

1. $f'(x) = 30x^5 - 90x^4 + 60x^3 = 0 \Rightarrow x = 0, 1, 2$.

2. $f''(x) = 150x^4 - 360x^3 + 180x^2$.

3. Higher Derivatives at $x=0$: f''(0) = 0, f'''(0) = 0, f^{iv}(0) = 360 > 0. Minimum at $x = 0, y_{min} = -10$.

4. At $x = 1$: f''(1) = -30 < 0. Max at $x = 1, y_{max} = -8$.

5. At $x = 2$: f''(2) = 240 > 0. Minimum at $x = 2, y_{min} = -26$.

Functions with No Maximum or Minimum Value

• Case (i): $x^3 - 3x^2 + 6x + 5$

  • $f'(x) = 3x^2 - 6x + 6 = 3[(x - 1)^2 + 1]$.

  • For any real values of $x$, $3[(x - 1)^2 + 1] \neq 0$.

  • Since $f'(x) \neq 0$, no maximum or minimum exists.

• Case (ii): $x^3 - 3x^2 + 3x + 1$

  • $f'(x) = 3x^2 - 6x + 3 = 3(x - 1)^2$.

  • At stationary point $x = 1$, $f''(1) = 6(1) - 6 = 0$.

  • $f'''(1) = 6 \neq 0$. Since the third derivative is non-zero, no relative maximum or minimum exists.

Point of Inflection

• Definition: If $f$ is continuous on an open interval containing $x_0$ and if $f$ changes the direction of its concavity at $x_0$, then the point $(x_0, f(x_0))$ is an inflection point.

• Condition: At an inflection point, $f''(x) = 0$.

• Example: For $y = x^3$, the point $x = 0$ is an inflection point.

Absolute Extremum

• Procedure: Evaluate $f$ at all critical points and at the endpoints $a$ and $b$. The largest value is the absolute maximum; the smallest is the absolute minimum.

• Sample Problem: Find absolute extrema of $f(x) = 2x^3 - 15x^2 + 36x$ on $[1, 5]$.

  • Critical points ($f'(x) = 0$): $6x^2 - 30x + 36 = 0 \Rightarrow x = 2, 3$.

  • Values:

    • $f(1) = 2(1)^3 - 15(1)^2 + 36(1) = 23$

    • $f(2) = 23(2)^3 - 15(2)^2 + 36(2) = 28$

    • $f(3) = 23(3)^3 - 15(3)^2 + 36(3) = 27$

    • $f(5) = 23(5)^3 - 15(5)^2 + 36(5) = 55$

  • Result: Absolute maximum is $55$ at $x=5$. Absolute minimum is $23$ at $x=1$.

Applications of Extrema

Problem 1: Minimum sum of a positive number and its reciprocal

1. Function: $y = x + \frac{1}{x}$.

2. Derivative: $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.

3. Setting to zero: $x^2 - 1 = 0 \Rightarrow x = 1$ (since x > 0).

4. Verification: x < 1 \Rightarrow y' = -; x > 1 \Rightarrow y' = +. Min at $x = 1$.

5. Value: $y = 1 + \frac{1}{1} = 2$.

Problem 2: Maximum volume of a folded box

1. Context: Sheet is $80\,cm \times 50\,cm$. Squares of side $x$ cut from corners.

2. Volume Formula: $y = (80 - 2x) \cdot (50 - 2x) \cdot x = 4000x - 260x^2 + 4x^3$.

3. Derivative: $\frac{dy}{dx} = 4000 - 520x + 12x^2 = 4(1000 - 130x + 3x^2)$.

4. Stationary point: $(3x - 100)(x - 10) = 0 \Rightarrow x = 10, \frac{100}{3}$.

5. Nature at $x = 10$: Small change analysis (x < 10 \Rightarrow +, x > 10 \Rightarrow -) shows maximum.

6. Max volume: At $x = 10$, $y = 18000\,cm^3$.

Problem 3: Minimum surface area of a box

1. Context: Length $3x$, width $x$, height $h$, volume $288\,cm^3$.

2. Height: $288 = 3x^2h \Rightarrow h = \frac{96}{x^2}$.

3. Surface area ($y$): $y = 2(3x \cdot x) + 2(xh) + 2(3xh) = 6x^2 + 8xh = 6x^2 + \frac{768}{x}$.

4. Derivative: $\frac{dy}{dx} = 12x - \frac{768}{x^2} = 0 \Rightarrow x^3 = 64 \Rightarrow x = 4$.

5. Nature: x < 4 \Rightarrow -; x > 4 \Rightarrow +. Minimum at $x = 4$.

6. Area value: $y = 6(16) + \frac{786}{4} = 288\,cm^2$.

Problem 4: Minimum material cost for a box

1. Context: Square base $x \, m$, height $h \, m$, volume $2000\,cm^3$. Top/bottom cost $3/cm^2$, sides cost $1.50/cm^2$.

2. Height: $h = \frac{2000}{x^2}$.

3. Cost: $y = (2x^2 \cdot 3) + (4xh \cdot 1.5) = 6x^2 + 6xh = 6x^2 + \frac{12000}{x}$.

4. Derivative: $\frac{dy}{dx} = 12x - \frac{12000}{x^2} = 0 \Rightarrow x = 10$.

5. Verification: Min at $x = 10$. Height $h = \frac{2000}{10^2} = 20\,m$.

6. Minimum cost: $y = 6(10^2) + \frac{12000}{10} = 1800$.