Calculus for Engineers: Derivatives, Extrema, and Mean Value Theorems

Definition and Interpretations of the Derivative

• Definition of the Derivative at a Point:   The derivative of a function $f$ at a point $x_0$, denoted $f'(x_0)$, is defined by the limit:   $f'(x_0) = \lim_{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}$   provided this limit exists.

• Interpretations of the Difference Quotient Limit:   The limit of the difference quotient, $\lim_{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}$, represents four equivalent mathematical and physical concepts:

  1. The slope of the graph of $y = f(x)$ at $x = x_0$

  2. The slope of the tangent line to the curve $y = f(x)$ at $x = x_0$

  3. The instantaneous rate of change of $f(x)$ with respect to $x$ at $x = x_0$

  4. The derivative value $f'(x_0)$ at a point

• Definition of the Derivative Function:   The derivative of the function $f(x)$ with respect to the variable $x$ is the function $f'$ whose value at $x$ is:   $f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$   provided the limit exists.

• Worked Example 1 — Differentiating via Definition:   Differentiate the function $f(x) = \frac{x}{x - 1}$.      Solution Steps:

  1. Evaluate $f(x + h)$:      $f(x + h) = \frac{x + h}{(x + h) - 1}$

  2. Apply the derivative definition:      $f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{\frac{x + h}{x + h - 1} - \frac{x}{x - 1}}{h}$

  3. Combine terms in the numerator using the algebraic identity $\frac{a}{b} - \frac{c}{d} = \frac{ad - cb}{bd}$:      $f'(x) = \lim_{h \rightarrow 0} \frac{1}{h} \cdot \frac{(x + h)(x - 1) - x(x + h - 1)}{(x + h - 1)(x - 1)}$

  4. Expand and simplify the numerator:      $(x + h)(x - 1) - x(x + h - 1) = (x^2 - x + hx - h) - (x^2 + xh - x) = -h$      $f'(x) = \lim_{h \rightarrow 0} \frac{1}{h} \cdot \frac{-h}{(x + h - 1)(x - 1)}$

  5. Cancel $h \neq 0$ and evaluate the limit as $h \rightarrow 0$:      $f'(x) = \lim_{h \rightarrow 0} \frac{-1}{(x + h - 1)(x - 1)} = \frac{-1}{(x - 1)^2}$

Absolute Extrema and Domain Restraints

• Definitions of Absolute Extrema:   Let $f$ be a function defined on domain $D$:

  • $f$ has an absolute maximum value on $D$ at a point $c$ if:     $f(x) \le f(c) \quad \text{for all } x \in D$

  • $f$ has an absolute minimum value on $D$ at $c$ if:     $f(x) \ge f(c) \quad \text{for all } x \in D$

• Comparative Examples with Function Rule $y = x^2$ Across Different Domains:

  • Case (a): Rule $y = x^2$, Domain $D = (-\infty, \infty)$

  • Absolute Maximum: None

  • Absolute Minimum: $0$ at $x = 0$

  • Case (b): Rule $y = x^2$, Domain $D = [0, 2]$

  • Absolute Maximum: $4$ at $x = 2$

  • Absolute Minimum: $0$ at $x = 0$

  • Case (c): Rule $y = x^2$, Domain $D = (0, 2]$

  • Absolute Maximum: $4$ at $x = 2$

  • Absolute Minimum: None

  • Case (d): Rule $y = x^2$, Domain $D = (0, 2)$

  • Absolute Maximum: None

  • Absolute Minimum: None

Extreme Value Theorem and Local Extrema

• Theorem 1 — The Extreme Value Theorem (EVT):   If $f$ is continuous on a closed interval $[a, b]$, then $f$ attains both an absolute maximum value $M$ and an absolute minimum value $m$ in $[a, b]$.   That is, there exist numbers $x_1$ and $x_2$ in $[a, b]$ such that:   $f(x_1) = m, \quad f(x_2) = M, \quad \text{and} \quad m \le f(x) \le M \quad \text{for every } x \in [a, b]$

• Definitions of Local Extrema:

  • A function $f$ has a local maximum value at a point $c$ within its domain $D$ if $f(x) \le f(c)$ for all $x \in D$ lying in some open interval containing $c$.

  • A function $f$ has a local minimum value at a point $c$ within its domain $D$ if $f(x) \ge f(c)$ for all $x \in D$ lying in some open interval containing $c$.

• Theorem 2 — The First Derivative Theorem for Local Extreme Values:   If $f$ has a local maximum or minimum value at an interior point $c$ of its domain, and if $f'$ is defined at $c$, then:   $f'(c) = 0$

• Definition of Critical Point:   An interior point of the domain of a function $f$ where $f'$ is zero or undefined is defined as a critical point of f$.\n\n# Finding Absolute Extrema on Closed Intervals\n\n- **Procedure to Find Absolute Extrema of Continuous Functions on Finite Closed Intervals:**\n 1. Evaluate f at all critical points and endpoints.\n 2. Compare all evaluated values: the largest value is the absolute maximum, and the smallest value is the absolute minimum.\n\n- **Worked Example 2:**\n  Find the absolute maximum and minimum values of f(x) = x^2$on$[-2, 1].\n  \n  *Solution Steps:*\n 1. Compute derivative: f'(x) = 2x.\n 2. Set f'(x) = 0 \implies 2x = 0 \implies x = 0$. The only critical point is$x = 0.\n 3. Evaluate f at critical point and endpoints:\n - Critical point value: f(0) = 0\n - Endpoint values: f(-2) = 4$,$f(1) = 1\n 4. Conclusion:\n - Absolute maximum value is 4$at$x = -2\n - Absolute minimum value is 0$at$x = 0\n\n- **Worked Example 3:**\n  Find the absolute maximum and minimum values of f(x) = 10x(2 - \ln(x))$on interval$[1, e^2].\n  \n  *Solution Steps:*\n 1. Calculate first derivative using product rule:\n     f'(x) = 10(2 - \ln(x)) - 10x \left(\frac{1}{x}\right) = 20 - 10\ln(x) - 10 = 10(1 - \ln(x))\n 2. Find critical points on [1, e^2]:\n     f'(x) = 0 \implies 10(1 - \ln(x)) = 0 \implies \ln(x) = 1 \implies x = e\n 3. Evaluate f$at critical point$x = e$and endpoints$x = 1$,$x = e^2:\n - Critical point value: f(e) = 10e(2 - \ln(e)) = 10e(2 - 1) = 10e \approx 27.2\n - Endpoint values:\n       f(1) = 10(1)(2 - \ln(1)) = 10(2 - 0) = 20\n       f(e^2) = 10e^2(2 - \ln(e^2)) = 10e^2(2 - 2) = 0\n 4. Conclusion:\n - Absolute maximum value is 10e \approx 27.2$at interior critical point$x = e\n - Absolute minimum value is 0$at right endpoint$x = e^2\n\n- **Worked Example 4:**\n  Find the absolute maximum and minimum values of f(x) = x^{2/3}$on interval$[-2, 3].\n  \n  *Solution Steps:*\n 1. Calculate first derivative:\n     f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}\n 2. f'(x)$has no zeros, but is undefined at interior point$x = 0$. Hence,$x = 0 is a critical point.\n 3. Evaluate f at critical point and endpoints:\n - Critical point value: f(0) = 0\n - Endpoint values:\n       f(-2) = (-2)^{2/3} = \sqrt[3]{4} \approx 1.587\n       f(3) = (3)^{2/3} = \sqrt[3]{9} \approx 2.08\n 4. Conclusion:\n - Absolute maximum value is \sqrt[3]{9} \approx 2.08$at right endpoint$x = 3\n - Absolute minimum value is 0$at interior point$x = 0 (where the graph exhibits a cusp)\n\n# Rolle's Theorem and the Mean Value Theorem\n\n- **Theorem 3 — Rolle's Theorem:**\n  Suppose that y = f(x)$is continuous over the closed interval$[a, b]$and differentiable at every point of its interior$(a, b)$. If$f(a) = f(b)$, then there is at least one number$c$in$(a, b) at which:\n  f'(c) = 0\n\n\n\n- **Worked Example 1 — Application of Rolle's Theorem:**\n  Show that the equation x^3 + 3x + 1 = 0 has exactly one real solution.\n  \n  *Solution Steps:*\n 1. Define continuous function f(x) = x^3 + 3x + 1.\n 2. Evaluate f(-1) = (-1)^3 + 3(-1) + 1 = -3$and$f(0) = 0^3 + 3(0) + 1 = 1.\n 3. By the Intermediate Value Theorem, because f(-1) < 0$and$f(0) > 0$, the graph of$f$crosses the x-axis at least once in the open interval$(-1, 0).\n 4. To prove uniqueness, assume for contradiction that there exist two roots x = a$and$x = b$where$f(a) = f(b) = 0.\n 5. By Rolle's Theorem, there would exist a point c \in (a, b)$such that$f'(c) = 0\n 6. Compute derivative: f'(x) = 3x^2 + 3$. Because$3x^2 + 3 \ge 3 > 0$for all real$x$,$f'(x) is never zero.\n 7. Since f'(x)$cannot be zero, a second root cannot exist. Therefore,$f(x) = 0 has exactly one real solution.\n\n- **Theorem 4 — The Mean Value Theorem (MVT):**\n  Suppose y = f(x)$is continuous over a closed interval$[a, b]$and differentiable on the interval's interior$(a, b)$. Then there is at least one point$c$in$(a, b) at which:\n  \frac{f(b) - f(a)}{b - a} = f'(c)\n\n- **Physical Interpretation of MVT:**\n  The quotient \frac{f(b) - f(a)}{b - a}$represents the average rate of change of$f$over$[a, b]$, and$f'(c) represents an instantaneous rate of change. The Mean Value Theorem asserts that at some interior point, the instantaneous rate of change equals the average rate of change across the entire interval.\n\n- **Corollaries of the Mean Value Theorem:**\n - **Corollary 1:** If f'(x) = 0$at each point$x$of an open interval$(a, b)$, then$f(x) = C$for all$x \in (a, b)$, where$C is a constant.\n - **Corollary 2:** If f'(x) = g'(x)$at each point$x$in an open interval$(a, b)$, then there exists a constant$C$such that$f(x) = g(x) + C$for all$x \in (a, b)$. That is,$f - g$is a constant function on$(a, b).\n\n- **Worked Example 4 — Finding Functions from Derivatives:**\n  Find the function f(x)$whose derivative is$\sin(x)$and whose graph passes through the point$(0, 2).\n  \n  *Solution Steps:*\n 1. Identify g(x) = -\cos(x)$, which has derivative$g'(x) = \sin(x).\n 2. Since f'(x) = g'(x) = \sin(x)$, Corollary 2 implies$f(x) = -\cos(x) + C$for some constant$C\n 3. Use point condition f(0) = 2:\n     f(0) = -\cos(0) + C = 2 \implies -1 + C = 2 \implies C = 3\n 4. Resulting function: f(x) = -\cos(x) + 3\n\n# Monotonicity and the First Derivative Test\n\n- **Corollary 3 — Increasing and Decreasing Function Criteria:**\n  Suppose that f$is continuous on$[a, b]$and differentiable on$(a, b).\n - If f'(x) > 0$at each point$x \in (a, b)$, then$f$is **increasing** on$[a, b].\n - If f'(x) < 0$at each point$x \in (a, b)$, then$f$is **decreasing** on$[a, b].\n\n- **Worked Example 1 — Intervals of Monotonicity:**\n  Find the critical points of f(x) = x^3 - 12x - 5$and identify open intervals on which$f is increasing and decreasing.\n  \n  *Solution Steps:*\n 1. Derivative: f'(x) = 3x^2 - 12 = 3(x^2 - 4) = 3(x + 2)(x - 2)\n 2. Critical points where f'(x) = 0$:$x = -2$and$x = 2\n 3. Subdivide real line into non-overlapping open intervals: (-\infty, -2)$,$(-2, 2)$, and$(2, \infty)\n 4. Evaluate sign of f' in each subinterval:\n - Interval (-\infty, -2)$: Evaluate$f'(-3) = 15 > 0$(Sign:$+$)$\implies f is increasing\n - Interval (-2, 2)$: Evaluate$f'(0) = -12 < 0$(Sign:$-$)$\implies f is decreasing\n - Interval (2, \infty)$: Evaluate$f'(3) = 15 > 0$(Sign:$+$)$\implies f is increasing\n\n- **First Derivative Test for Local Extrema:**\n  Suppose c$is a critical point of a continuous function$f$, and$f$is differentiable at every point in some interval containing$c$except possibly at$c itself. Moving across this interval from left to right:\n 1. If f'$changes from negative to positive at$c$, then$f$has a **local minimum** at$c.\n 2. If f'$changes from positive to negative at$c$, then$f$has a **local maximum** at$c.\n 3. If f'$does not change sign at$c$($f'$is positive on both sides or negative on both sides), then$f$has **no local extremum** at$c$.

• Worked Example 2 — Applying First Derivative Test:   Find critical points of $f(x) = x^{1/3}(x - 4) = x^{4/3} - 4x^{1/3}$. Identify open intervals of increasing/decreasing, and find local and absolute extreme values.      Solution Steps:

  1. $f$ is continuous everywhere as a product of continuous functions.

  2. Compute derivative:      $f'(x) = \frac{d}{dx}\left(x^{4/3} - 4x^{1/3}\right) = \frac{4}{3}x^{1/3} - \frac{4}{3}x^{-2/3} = \frac{4}{3}x^{-2/3}(x - 1) = \frac{4(x - 1)}{3x^{2/3}}$

  3. Identify critical points:

  • $f'(x) = 0 \implies x = 1$

  • $f'(x)$ is undefined at $x = 0$

  • Critical points: $x = 0$ and $x = 1$

  1. Test subinterval signs:

  • Interval $x < 0$: Sign of $f'$ is $-$ $\implies f$ is decreasing

  • Interval $0 < x < 1$: Sign of $f'$ is $-$ $\implies f$ is decreasing

  • Interval $x > 1$: Sign of $f'$ is $+$ $\implies f$ is increasing

  1. Extreme values analysis:

  • At $x = 0$: $f'$ does not change sign (negative on both sides), so no local extremum at $x = 0$.

  • At $x = 1$: $f'$ changes from negative to positive, so $f$ has a local minimum at $x = 1$. Value: $f(1) = 1^{1/3}(1 - 4) = -3$.

• Worked Example 3 — Comprehensive Monotonicity and Extrema Analysis:   Find critical points of $f(x) = (x^2 - 3)e^x$. Identify open intervals of increasing/decreasing, and find local and absolute extreme values.      Solution Steps:

  1. Function $f$ is continuous and differentiable for all real numbers.

  2. Compute derivative using product rule:      $f'(x) = (x^2 - 3)\frac{d}{dx}(e^x) + \frac{d}{dx}(x^2 - 3) e^x = (x^2 - 3)e^x + (2x)e^x = (x^2 + 2x - 3)e^x$

  3. Set $f'(x) = 0$. Since $e^x > 0$ for all $x$:      $x^2 + 2x - 3 = 0 \implies (x + 3)(x - 1) = 0$      Critical points are $x = -3$ and $x = 1$

  4. Partition real line and test signs:

  • Interval $x < -3$: Sign of $f'$ is $+$ $\implies f$ is increasing

  • Interval $-3 < x < 1$: Sign of $f'$ is $-$ $\implies f$ is decreasing

  • Interval $1 < x$: Sign of $f'$ is $+$ $\implies f$ is increasing

  1. Determine local and absolute extrema:

  • Local Maximum at $x = -3$: Value $f(-3) = ((-3)^2 - 3)e^{-3} = 6e^{-3} \approx 0.299$

  • Local Minimum at $x = 1$: Value $f(1) = (1^2 - 3)e^1 = -2e \approx -5.437$

  • Absolute Extrema: The local minimum value $-2e \approx -5.437$ at $x = 1$ is also an absolute minimum because $f(x) > 0$ for $|x| > \sqrt{3}$. There is no absolute maximum since $\lim_{x \to \infty} f(x) = \infty$.