Applications of Derivatives

A critical point of $f$ occurs at $x=c$ where $f'(c) = 0$ or $f'(c)$ is undefined (and $f(c)$ exists).

The first derivative test: examine the sign of $f'(x)$ on either side of the critical point.

If $f'$ changes from $+$ to $-$ at $c$, then $f(c)$ is a local maximum.

If $f'$ changes from $-$ to $+$ at $c$, then $f(c)$ is a local minimum.

If $f'$ does not change sign (e.g., $f(x) = x^3$ at $x=0$), the critical point is neither a max nor a min.

To apply: find all critical points, then build a sign chart for $f'(x)$ using test values in each interval.

An alternative to sign charts: at a critical point where $f'(c) = 0$:

If $f''(c) > 0$, the function is concave up at $c$, so $f(c)$ is a local minimum (think: bowl shape).

If $f''(c) < 0$, the function is concave down at $c$, so $f(c)$ is a local maximum (think: arch shape).

If $f''(c) = 0$, the test is inconclusive. Fall back to the first derivative test.

The second derivative test is often faster when $f''$ is easy to compute.

Concavity describes how the curve bends. $f''(x) > 0$ means concave up (opening upward), $f''(x) < 0$ means concave down (opening downward).

An inflection point is where the concavity changes: $f''$ switches sign. At an inflection point, the curve transitions from bending one way to bending the other.

To find inflection points: set $f''(x) = 0$ (or find where $f''$ is undefined), then verify $f''$ changes sign across that point.

Example: $f(x) = x^3$ has $f''(x) = 6x$. This equals $0$ at $x=0$, and changes from negative to positive. So $(0, 0)$ is an inflection point.

Note: $f''(c) = 0$ alone is not enough. $f(x) = x^4$ has $f''(0) = 0$ but no inflection point (concavity doesn't change).

The Extreme Value Theorem: if $f$ is continuous on a closed interval $[a,b]$, then $f$ attains an absolute maximum and an absolute minimum on $[a,b]$.

The closed interval method to find them:

Step 1: Find all critical points of $f$ in the open interval $(a,b)$.

Step 2: Evaluate $f$ at each critical point and at both endpoints $a$ and $b$.

Step 3: The largest value is the absolute maximum, the smallest is the absolute minimum.

This is the go-to method for any 'find the max/min on an interval' problem. Never forget the endpoints!

A complete curve sketch uses the following information from derivatives:

1. Domain and intercepts: where is $f$ defined? Where does it cross the axes?

2. First derivative analysis: find critical points, determine intervals of increase/decrease.

3. Second derivative analysis: find inflection points, determine concavity on each interval.

4. Asymptotes: check for vertical asymptotes (denominator = 0) and horizontal asymptotes ($\lim_{x\to\pm\infty}$).

5. Plot key points (critical points, inflection points, intercepts) and connect with the correct shape (increasing/decreasing, concave up/down).

Example: for $f(x) = x^3 - 3x$: $f'(x) = 3x^2-3 = 3(x-1)(x+1)$. Critical points at $x=\pm 1$. $f''(x) = 6x$, inflection point at $x=0$. Local max at $(-1, 2)$, local min at $(1, -2)$, inflection at $(0, 0)$.

The general strategy for optimization word problems:

Step 1: Draw a diagram and define variables. Label everything.

Step 2: Write the objective function (what you want to maximize or minimize).

Step 3: Write the constraint equation (the relationship that limits your variables).

Step 4: Use the constraint to eliminate one variable from the objective, getting a function of a single variable.

Step 5: Find the critical points of this function. Test them (and the endpoints, if the domain is closed) to identify the optimum.

Step 6: Answer the original question. Make sure you're solving for what was asked.

Common setups: maximize area given a perimeter constraint, minimize material given a volume constraint, maximize revenue.

Related rates problems involve multiple quantities changing over time, connected by a geometric or physical equation.

Step 1: Draw a picture and identify all variables. Label which quantities are changing.

Step 2: Write an equation relating the variables (e.g., Pythagorean theorem, area formula, volume formula).

Step 3: Differentiate both sides with respect to time $t$ using implicit differentiation.

Step 4: Plug in all known values and rates at the specific instant in question.

Step 5: Solve for the unknown rate.

Critical: never plug in specific values before differentiating. The equation must remain general during differentiation because the variables are changing.

Common setups: expanding balloon (sphere volume), sliding ladder (Pythagorean theorem), filling cone (similar triangles + cone volume), spreading oil slick (circle area).

Given a position function $s(t)$: velocity is $v(t) = s'(t)$ and acceleration is $a(t) = v'(t) = s''(t)$.

The particle is moving right when $v(t) > 0$ and moving left when $v(t) < 0$.

The particle stops (changes direction) when $v(t) = 0$ and the velocity changes sign.

Speed is $|v(t)|$. The particle speeds up when velocity and acceleration have the same sign, and slows down when they have opposite signs.

Total distance traveled is $\int_a^b |v(t)|\,dt$ (not the same as displacement $\int_a^b v(t)\,dt$).

The linearization of $f$ at $a$ is $L(x) = f(a) + f'(a)(x-a)$. It's the tangent line used as an approximation.

For values of $x$ near $a$, $f(x) \approx L(x)$. This is how engineers and scientists do quick estimates.

Differentials: if $y = f(x)$, then $dy = f'(x)\,dx$. The differential $dy$ approximates the actual change $\Delta y$.

Example: estimate $\sqrt{4.02}$. Use $f(x) = \sqrt{x}$ at $a=4$. $f'(4) = 1/4$. So $\sqrt{4.02} \approx 2 + \frac{1}{4}(0.02) = 2.005$.

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one $c$ in $(a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.

In words: somewhere between $a$ and $b$, the instantaneous rate of change equals the average rate of change.

Geometric interpretation: there's a point where the tangent line is parallel to the secant line connecting $(a, f(a))$ and $(b, f(b))$.

Applications: proving that a function must have a certain derivative value, establishing speed limits (if you drove 120 miles in 2 hours, you must have been going 60 mph at some instant).

Newton's method finds approximate roots of $f(x) = 0$ using tangent lines.

Start with an initial guess $x_0$. The iteration formula is: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$.

Each step draws the tangent line at $(x_n, f(x_n))$ and finds where it crosses the $x$-axis. That crossing point is $x_{n+1}$.

Convergence is typically very fast (quadratic) when the initial guess is close to the root.

Pitfalls: the method can fail if $f'(x_n) = 0$ (horizontal tangent), if the initial guess is too far from the root, or if the function oscillates.