Critical Points, Derivative Tests, and Applications in Calculus

Critical Points and Intervals of Increase/Decrease

Identifying Increasing and Decreasing Intervals

Understanding where a function increases or decreases is fundamental in calculus. This involves analyzing the first derivative and interpreting its sign.

• Find the first derivative $f'(x)$ of the function $f(x)$.

• Identify critical points by solving $f'(x) = 0$ and finding where $f'(x)$ is undefined.

• Create a sign chart for $f'(x)$ to determine the sign (positive or negative) in each interval between critical points.

• Write intervals of increase (where $f'(x) > 0$) and decrease (where $f'(x) < 0$) using proper interval notation.

Example: For $f(x) = x^3 - 3x^2 + 2$, $f'(x) = 3x^2 - 6x$. Set $f'(x) = 0$ to find critical points at $x = 0$ and $x = 2$. Test intervals to determine where $f(x)$ increases or decreases.

First and Second Derivative Tests

First Derivative Test

The first derivative test uses the sign changes of $f'(x)$ around critical points to classify local extrema.

• If $f'(x)$ changes from positive to negative at $x = c$, $f$ has a local maximum at $c$.

• If $f'(x)$ changes from negative to positive at $x = c$, $f$ has a local minimum at $c$.

• If there is no sign change, $c$ is neither a maximum nor a minimum.

Example: For $f(x) = x^3$, $f'(x) = 3x^2$; at $x = 0$, there is no sign change, so no local extremum.

Second Derivative Test

The second derivative test provides another method for classifying critical points using $f''(x)$.

• Find $f''(x)$ and evaluate at the critical point $c$.

• If $f''(c) > 0$, $f$ has a local minimum at $c$.

• If $f''(c) < 0$, $f$ has a local maximum at $c$.

• If $f''(c) = 0$, the test is inconclusive.

Example: For $f(x) = x^2$, $f'(x) = 2x$, $f''(x) = 2$. At $x = 0$, $f''(0) = 2 > 0$, so there is a local minimum.

Absolute Maximum and Minimum (Extreme Value Theorem)

Finding Absolute Extrema on a Closed Interval

The Extreme Value Theorem guarantees that a continuous function on a closed interval $[a, b]$ attains both an absolute maximum and minimum.

• Find all critical points inside the interval.

• Evaluate $f(x)$ at all critical points and at the endpoints $a$ and $b$.

• Compare all values to determine the absolute maximum and minimum.

Example: For $f(x) = x^2$ on $[-1, 2]$, check $f(-1)$, $f(0)$, and $f(2)$ to find the absolute minimum and maximum.

Differentials and Linear Approximation

Using Differentials for Approximation

Differentials provide a method for approximating small changes in a function's value.

• The differential is defined as $dy = f'(x)dx$.

• Given a small change $dx$, $dy$ approximates the change in $f(x)$.

• This method gives an approximation, not the exact value.

Example: To approximate $\sqrt{4.1}$, let $f(x) = \sqrt{x}$, $x = 4$, $dx = 0.1$, $f'(x) = \frac{1}{2\sqrt{x}}$, so $dy = \frac{1}{4} \cdot 0.1 = 0.025$; thus, $\sqrt{4.1} \approx 2 + 0.025 = 2.025$.

Exponential Growth and Decay (Half-Life)

Solving Exponential Growth and Decay Problems

Exponential growth and decay are modeled by differential equations involving a constant rate of change proportional to the current amount.

• The general solution is $y(t) = y_0 e^{kt}$, where $k$ is the growth ($k > 0$) or decay ($k < 0$) constant.

• To find $k$, use given information and solve for $k$ using natural logarithms.

• For half-life problems, set $y(t_{1/2}) = \frac{1}{2}y_0$ and solve for $t_{1/2}$.

Example: If a substance has a half-life of 10 years, $\frac{1}{2} = e^{10k} \implies k = \frac{\ln(1/2)}{10}$.

Implicit Differentiation

Differentiating Equations Involving $y$

Implicit differentiation is used when $y$ is defined implicitly as a function of $x$.

• Differentiate both sides of the equation with respect to $x$.

• Apply the chain rule to terms involving $y$ (i.e., $\frac{d}{dx}y = y'$).

• Solve for $y'$ (or $\frac{dy}{dx}$).

Example: For $x^2 + y^2 = 1$, differentiating both sides gives $2x + 2y y' = 0 \implies y' = -\frac{x}{y}$.

Derivative Rules (All Combined)

Applying All Differentiation Rules

Mastery of all derivative rules is essential, especially for complex functions involving combinations of rules.

• Power Rule: $\frac{d}{dx}x^n = nx^{n-1}$

• Product Rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$

• Quotient Rule: $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

• Chain Rule: $\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$

• Exponential Derivatives: $\frac{d}{dx}e^{u(x)} = e^{u(x)}u'(x)$

• Logarithmic Derivatives: $\frac{d}{dx}\ln(u(x)) = \frac{u'(x)}{u(x)}$

Example: For $f(x) = e^{2x^2 + 3x}$, $f'(x) = e^{2x^2 + 3x} \cdot (4x + 3)$ (chain rule applied).

Logarithmic Rules

Properties and Manipulation of Logarithms

Logarithmic properties are essential for simplifying expressions and solving equations involving logarithms.

• Product Rule: $\ln(ab) = \ln a + \ln b$

• Quotient Rule: $\ln\left(\frac{a}{b}\right) = \ln a - \ln b$

• Power Rule: $\ln(a^k) = k\ln a$

• Be able to expand and combine logarithmic expressions using these properties.

Example: $\ln(2x^3) = \ln 2 + 3\ln x$

Concavity and Inflection Points

Determining Concavity

Concavity describes the direction a curve bends. It is determined by the sign of the second derivative.

• If $f''(x) > 0$, $f$ is concave up on that interval.

• If $f''(x) < 0$, $f$ is concave down on that interval.

• Write intervals of concavity using proper notation.

Example: For $f(x) = x^4$, $f''(x) = 12x^2$, which is always $\geq 0$, so $f$ is concave up everywhere.

Inflection Points

Inflection points occur where the function changes concavity.

• Find where $f''(x) = 0$ or is undefined.

• Confirm that $f''(x)$ changes sign at these points.

• State inflection points as ordered pairs $(x, f(x))$.

Example: For $f(x) = x^3$, $f''(x) = 6x$. At $x = 0$, $f''(x)$ changes sign, so $(0, 0)$ is an inflection point.

Absolute Extrema Using the Closed Interval Method

Applying the Closed Interval Method

To find absolute maximum and minimum values on a closed interval, use the closed interval method:

• Find all critical points inside the interval.

• Evaluate the function at all critical points and endpoints.

• Compare all values to determine the absolute maximum and minimum.

Example: For $f(x) = -x^2 + 4x$ on $[0, 5]$, check $f(0)$, $f(2)$, and $f(5)$.

Optimization Problems

Solving Optimization Problems

Optimization involves finding the maximum or minimum value of a function under given constraints.

1. Define variables clearly.

2. Write the objective function to be maximized or minimized.

3. Use the constraint to reduce the objective function to one variable.

4. Take the derivative and find critical points.

5. Answer the question clearly, including units if applicable.

Example: Find the dimensions of a rectangle with perimeter 20 that maximize area. Let $x$ and $y$ be sides, $2x + 2y = 20 \implies y = 10 - x$. Area $A = x(10 - x)$. Maximize $A$ by finding $A'(x) = 10 - 2x = 0 \implies x = 5$, so $y = 5$.