Advanced Calculus Comprehensive Final Review Notes and Review

Analysis of Power Functions with Fractional Exponents

This section focuses on the behavior of the function $f(x) = (x - 3)^{1/5}$ through the lens of first and second derivatives.

First Derivative Analysis: Intervals of Increase and Decrease

• Definition of the First Derivative: To find where the function is increasing or decreasing, we must first calculate $f'(x)$.     $f'(x) = \frac{1}{5}(x - 3)^{-4/5} = \frac{1}{5(x - 3)^{4/5}}$
• Critical Points: Setting the numerator to zero yields no solutions. The derivative $f'(x)$ is undefined at $x = 3$ because the denominator becomes zero. Thus, $x = 3$ is the critical value.
• Interval Testing for $f'(x)$:
  • For the interval $(-\infty, 3)$: Choose $x = 2$. $f'(2) = \frac{1}{5(2 - 3)^{4/5}} = \frac{1}{5(-1)^{4/5}}$. Since $(-1)^4 = 1$ and the fifth root of $1$ is $1$, $f'(2) = \frac{1}{5}$, which is positive.
  • For the interval $(3, \infty)$: Choose $x = 4$. $f'(4) = \frac{1}{5(4 - 3)^{4/5}} = \frac{1}{5(1)^{4/5}} = \frac{1}{5}$, which is positive.
• Justification: Because $f'(x) > 0$ for all $x \neq 3$ and the function is continuous at $x = 3$, $f(x)$ is increasing on the entire domain $(-\infty, \infty)$.

Relative Extrema

• Local Maximum and Minimum Values: A relative maximum or minimum occurs only if $f'(x)$ changes sign at a critical point.
• Justification: As demonstrated in the interval testing above, $f'(x)$ remains positive on both sides of the critical point $x = 3$. Therefore, there are no local maximum or minimum values for this function.

Second Derivative Analysis: Concavity and Inflection

• Calculating the Second Derivative: To determine concavity, we differentiate $f'(x)$.     $f''(x) = \frac{d}{dx} \left[ \frac{1}{5}(x - 3)^{-4/5} \right] = \frac{1}{5} \times \left( -\frac{4}{5} \right) (x - 3)^{-9/5} = -\frac{4}{25(x - 3)^{9/5}}$
• Concavity Intervals:
  • Concave Up: The function is concave up where $f''(x) > 0$. Testing $x < 3$: $(x - 3)$ is negative. A negative number raised to an odd power ($9$) remains negative, and the fifth root remains negative. The expression $-\frac{4}{25 \times (\text{negative})}$ results in a positive value. Thus, $f(x)$ is concave up on $(-\infty, 3)$.
  • Concave Down: The function is concave down where $f''(x) < 0$. Testing $x > 3$: $(x - 3)$ is positive, making $f''(x)$ negative due to the leading negative coefficient. Thus, $f(x)$ is concave down on $(3, \infty)$.
• Inflection Points: An inflection point exists where the concavity changes and the function is defined.
• Justification: Since the concavity changes from concave up to concave down at $x = 3$, there is an inflection point at $x = 3$.

Analyzing the Graph of the Derivative function $f'(x)$

When provided with the graph of $f'$, the following principles are used to describe the behavior of the original function $f$.

Identifying Relative Extrema from $f'$

• Relative Minimum: This occurs at values of $x$ where $f'$ changes from negative to positive (crosses the x-axis from below to above).
• Relative Maximum: This occurs at values of $x$ where $f'$ changes from positive to negative (crosses the x-axis from above to below).
• Justification: The sign of $f'$ indicates the slope of $f$; a change from negative to positive slope indicates a valley (minimum), while positive to negative indicates a peak (maximum).

Identifying Points of Inflection from $f'$

• Point of Inflection: These occur at the $x$-values where the graph of $f'$ has a relative maximum or relative minimum.
• Justification: A relative extreme on $f'$ means the slope of $f'$ (which is $f''$) changes sign. A sign change in $f''$ defines a point of inflection for $f$.

Intervals of Increase and Specific Concavity

• Increasing and Concave Down: To satisfy both conditions simultaneously:
  1. The function $f$ is increasing where $f'(x) > 0$ (the graph of $f'$ is above the x-axis).
  2. The function $f$ is concave down where $f'(x)$ is decreasing (the slope of $f'$ is negative).
• Justification: The interval(s) required are the regions where the $f'$ graph is both positive and trending downwards.

Differentiability and Continuity Requirements

To find missing constants that make a piecewise function differentiable, two conditions must be met at the point of interest, $x = c$.

1. Continuity: The limit from the left must equal the limit from the right at $x = c$.     $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$
2. Differentiability: The derivative from the left must equal the derivative from the right at $x = c$.     $\lim_{x \to c^-} f'(x) = \lim_{x \to c^+} f'(x)$

Derivative Rules and Table-Based Calculations

The following table of values is used for calculations at specific points:

Problem 4: Product Rule Application

• Function: $h(x) = f(x)g(x)$
• Derivative Rule: $h'(x) = f'(x)g(x) + f(x)g'(x)$
• Calculation for $h'(2)$:     $h'(2) = f'(2)g(2) + f(2)g'(2)$$h'(2) = (-2)(5) + (3)(4)$$h'(2) = -10 + 12 = 2$

Problem 5: Chain Rule Application

• Function: $k(x) = f(g(x))$
• Derivative Rule: $k'(x) = f'(g(x)) \times g'(x)$
• Calculation for $k'(2)$:     $k'(2) = f'(g(2)) \times g'(2)$$k'(2) = f'(5) \times 4$$k'(2) = (11) \times 4 = 44$

Problem 6: Quotient Rule Application

• Function: $q(x) = \frac{f(x)}{g(x)}$
• Derivative Rule: $q'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$
• Calculation for $q'(5)$:     $q'(5) = \frac{f'(5)g(5) - f(5)g'(5)}{(g(5))^2}$$q'(5) = \frac{(11)(-6) - (9)(7)}{(-6)^2}$$q'(5) = \frac{-66 - 63}{36} = \frac{-129}{36} = -\frac{43}{12}$

General Derivative Expressions in Terms of $g(x)$

Finding $f'(x)$ when $f(x)$ is defined as a composition or product involving an unknown function $g(x)$.

• Problem 7: $f(x) = x^3 \cdot g(x)$     Using the Product Rule:     $f'(x) = 3x^2 g(x) + x^3 g'(x)$

• Problem 8: $f(x) = [g(x)]^3$     Using the Power Chain Rule:     $f'(x) = 3[g(x)]^2 \times g'(x)$

• Problem 9: $f(x) = g(x^3)$     Using the Chain Rule:     $f'(x) = g'(x^3) \times 3x^2$

• Problem 10: $f(x) = [g(x^3)]^3$     Using the Generalized Chain Rule:     $f'(x) = 3[g(x^3)]^2 \times g'(x^3) \times 3x^2 = 9x^2 [g(x^3)]^2 g'(x^3)$

Implicit Differentiation and Tangent Lines

Finding $\frac{dy}{dx}$ and Tangent Line Equations

• Equation: $x^4 y = 3y - x$
• Implicit Differentiation (Differentiating both sides with respect to $x$):     $\frac{d}{dx}(x^4 y) = \frac{d}{dx}(3y - x)$$4x^3 y + x^4 \frac{dy}{dx} = 3 \frac{dy}{dx} - 1$
• Solving for $\frac{dy}{dx}$:     $x^4 \frac{dy}{dx} - 3 \frac{dy}{dx} = -1 - 4x^3 y$$\frac{dy}{dx}(x^4 - 3) = -(1 + 4x^3 y)$$\frac{dy}{dx} = \frac{-(1 + 4x^3 y)}{x^4 - 3}$
• Evaluating at the point $(-1, -1)$:     $\text{Slope } m = \frac{-(1 + 4(-1)^3(-1))}{(-1)^4 - 3} = \frac{-(1 + 4)}{1 - 3} = \frac{-5}{-2} = \frac{5}{2}$
• Equation of the Tangent Line:     Using $y - y_1 = m(x - x_1)$:     $y - (-1) = \frac{5}{2}(x - (-1))$$y + 1 = \frac{5}{2}(x + 1)$$y = \frac{5}{2}x + \frac{5}{2} - 1 \implies y = \frac{5}{2}x + \frac{3}{2}$

Analysis of the Curve $4x^2 - 24x + y^2 + 4y = -36$

Part A: Expression for the Slope

• Differentiation:     $8x - 24 + 2y \frac{dy}{dx} + 4 \frac{dy}{dx} = 0$
• Isolating $\frac{dy}{dx}$:     $(2y + 4) \frac{dy}{dx} = 24 - 8x$$\frac{dy}{dx} = \frac{24 - 8x}{2y + 4} = \frac{8(3 - x)}{2(y + 2)} = \frac{4(3 - x)}{y + 2}$

Part B: Horizontal Tangents

• Condition: A horizontal tangent occurs when the numerator of the slope is zero and the denominator is non-zero.     $4(3 - x) = 0 \implies x = 3$
• Finding Coordinates: Plug $x = 3$ into the original equation:     $4(3)^2 - 24(3) + y^2 + 4y = -36$$36 - 72 + y^2 + 4y = -36$$-36 + y^2 + 4y = -36$$y^2 + 4y = 0 \implies y(y + 4) = 0$
• Resulting Points: $(3, 0)$ and $(3, -4)$.

Part C: Vertical Tangents

• Condition: A vertical tangent occurs when the denominator of the slope is zero and the numerator is non-zero.     $y + 2 = 0 \implies y = -2$
• Finding Coordinates: Plug $y = -2$ into the original equation:     $4x^2 - 24x + (-2)^2 + 4(-2) = -36$$4x^2 - 24x + 4 - 8 = -36$$4x^2 - 24x - 4 = -36$$4x^2 - 24x + 32 = 0$     Divide by $4$:     $x^2 - 6x + 8 = 0 \implies (x - 4)(x - 2) = 0$
• Resulting Points: $(4, -2)$ and $(2, -2)$.