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Applications of Derivatives

Derivatives are powerful tools that allow us to analyze and understand functions in a deeper way. They help us determine how a function behaves, locate critical points, optimize real-world problems, and even approximate complex functions. Below, we explore several key applications of derivatives.

1. Tangent Lines and Rates of Change

The derivative \( f'(x) \) gives the slope of the tangent line at a given point \( x \). This means that the equation of the tangent line to the function \( f(x) \) at \( x = a \) is given by:

\[ y - f(a) = f'(a)(x - a) \]

Example 1: Finding a Tangent Line

Find the equation of the tangent line to \( f(x) = x^3 - 2x + 1 \) at \( x = 1 \).

Step 1: Compute \( f(1) \):

\[ f(1) = (1)^3 - 2(1) + 1 = 0 \]

Step 2: Compute \( f'(x) \):

\[ f'(x) = 3x^2 - 2 \]

Step 3: Evaluate \( f'(1) \):

\[ f'(1) = 3(1)^2 - 2 = 1 \]

Step 4: Use the tangent line equation:

\[ y - 0 = 1(x - 1) \] \[ y = x - 1 \]

2. Increasing and Decreasing Functions

A function \( f(x) \) is:

  • Increasing if \( f'(x) > 0 \) for all \( x \) in an interval.
  • Decreasing if \( f'(x) < 0 \) for all \( x \) in an interval.

The first derivative test helps determine whether a function is increasing or decreasing.

Example 2: Finding Intervals of Increase and Decrease

Determine where \( f(x) = x^3 - 3x^2 + 4 \) is increasing or decreasing.

Step 1: Compute the derivative:

\[ f'(x) = 3x^2 - 6x \]

Step 2: Find critical points by solving \( f'(x) = 0 \):

\[ 3x^2 - 6x = 0 \] \[ 3x(x - 2) = 0 \] \[ x = 0, 2 \]

Step 3: Test sign changes in intervals \( (-\infty, 0) \), \( (0, 2) \), and \( (2, \infty) \).

  • For \( x = -1 \), \( f'(-1) = 3(1) + 6(1) = 9 > 0 \) → Increasing
  • For \( x = 1 \), \( f'(1) = 3(1) - 6(1) = -3 < 0 \) → Decreasing
  • For \( x = 3 \), \( f'(3) = 3(9) - 6(3) = 9 > 0 \) → Increasing

Conclusion: \( f(x) \) is increasing on \( (-\infty, 0) \cup (2, \infty) \) and decreasing on \( (0, 2) \).

3. Local Extrema and Critical Points

A function has a local maximum at \( x = c \) if \( f(c) \) is larger than nearby values, and a local minimum if \( f(c) \) is smaller. These occur at critical points, where \( f'(x) = 0 \) or is undefined.

4. Optimization Problems

Derivatives help solve real-world optimization problems, such as maximizing profit or minimizing cost.

Example 3: Maximizing Area

A farmer wants to enclose a rectangular field with 200 meters of fencing. What dimensions maximize the area?

Step 1: Define variables:

\[ \text{Let width } x, \text{ then length } y = \frac{200 - 2x}{2} = 100 - x \]

Step 2: Express area:

\[ A = x(100 - x) \]

Step 3: Differentiate:

\[ A' = 100 - 2x \]

Step 4: Solve \( A' = 0 \):

\[ 100 - 2x = 0 \] \[ x = 50 \]

Since \( A''(x) = -2 \) is negative, \( x = 50 \) is a maximum. So, the optimal dimensions are 50m × 50m.

Practice Problems

Problem 1: Find the equation of the tangent line to \( f(x) = x^2 - 4x + 3 \) at \( x = 2 \).

Solution:

Compute \( f'(x) = 2x - 4 \), then evaluate at \( x = 2 \):

\[ f'(2) = 2(2) - 4 = 0 \]

Find \( f(2) \):

\[ f(2) = (2)^2 - 4(2) + 3 = 3 \]

Use tangent line formula:

\[ y - 3 = 0(x - 2) \Rightarrow y = 3 \]

Problem 2: Find the critical points of \( f(x) = x^3 - 6x^2 + 9x + 1 \) and classify them.

Solution:

Compute \( f'(x) \):

\[ f'(x) = 3x^2 - 12x + 9 \]

Find where \( f'(x) = 0 \):

\[ 3x^2 - 12x + 9 = 0 \] \[ (x - 3)(x - 1) = 0 \] \[ x = 1, 3 \]

Using the second derivative test, \( f''(x) = 6x - 12 \).

  • At \( x = 1 \), \( f''(1) = -6 \) (concave down) → local max
  • At \( x = 3 \), \( f''(3) = 6 \) (concave up) → local min

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