Function Analysis with Derivatives

Derivatives provide a powerful way to understand the behavior of functions. By analyzing the first and second derivatives, we can sketch highly accurate graphs and identify key features like where a function is increasing, decreasing, curving upwards, or curving downwards.

The First Derivative: Increasing, Decreasing, and Extrema

The first derivative, $f'(x)$, tells us the slope of the tangent line to the function.

• Increasing: A function is increasing on an interval if $f'(x) > 0$.
• Decreasing: A function is decreasing on an interval if $f'(x) < 0$.
• Critical Points: These occur where $f'(x) = 0$ or $f'(x)$ is undefined. They are the "candidates" for local maximums and minimums.

The First Derivative Test states that if $f'(x)$ changes from positive to negative at a critical point, it is a local maximum. If it changes from negative to positive, it is a local minimum.

Example 1: Find local extrema and intervals of increase/decrease

Let $f(x) = x^3 - 12x + 1$.

1. Find the derivative: $f'(x) = 3x^2 - 12$
2. Find critical points: Set $f'(x) = 0$. $3(x^2 - 4) = 0 \implies x = 2, x = -2$
3. Test intervals:
   • For $x < -2$ (e.g., $x = -3$): $f'(-3) = 15 > 0$ (Increasing)
   • For $-2 < x < 2$ (e.g., $x = 0$): $f'(0) = -12 < 0$ (Decreasing)
   • For $x > 2$ (e.g., $x = 3$): $f'(3) = 15 > 0$ (Increasing)

Conclusion:

• Intervals of increase: $(-\infty, -2) \cup (2, \infty)$
• Interval of decrease: $(-2, 2)$
• Local maximum at $x = -2$ (value: $f(-2) = 17$)
• Local minimum at $x = 2$ (value: $f(2) = -15$)

The Second Derivative: Concavity and Inflection Points

The second derivative, $f''(x)$, tells us the rate of change of the first derivative. It describes the concavity of the function.

• Concave Up: If $f''(x) > 0$, the graph is shaped like a cup ($\cup$).
• Concave Down: If $f''(x) < 0$, the graph is shaped like a frown ($\cap$).
• Inflection Points: A point where the concavity changes (from up to down, or down to up). This occurs where $f''(x) = 0$ or is undefined, and $f''(x)$ changes sign.

Example 2: Find the inflection points

Let $f(x) = x^4 - 4x^3$.

1. Find the first and second derivatives: $f'(x) = 4x^3 - 12x^2$ $f''(x) = 12x^2 - 24x$
2. Find potential inflection points: Set $f''(x) = 0$. $12x(x - 2) = 0 \implies x = 0, x = 2$
3. Test intervals for concavity:
   • For $x < 0$ (e.g., $x = -1$): $f''(-1) = 36 > 0$ (Concave up)
   • For $0 < x < 2$ (e.g., $x = 1$): $f''(1) = -12 < 0$ (Concave down)
   • For $x > 2$ (e.g., $x = 3$): $f''(3) = 36 > 0$ (Concave up)

Conclusion: Since the concavity changes at both $x = 0$ and $x = 2$, both are inflection points.

• At $x = 0$, the point is $(0, 0)$.
• At $x = 2$, the point is $(2, -16)$.

Absolute Extrema and the Extreme Value Theorem

The Extreme Value Theorem (EVT) guarantees that if a function is continuous on a closed interval $[a, b]$, it must have both an absolute maximum and an absolute minimum on that interval.

To find these absolute extrema:

1. Find all critical points within the open interval $(a, b)$.
2. Evaluate the function $f(x)$ at these critical points.
3. Evaluate the function at the endpoints $x = a$ and $x = b$.
4. The largest value is the absolute maximum, and the smallest is the absolute minimum.