Problem 9 - Calculus: Early Transcendentals

Question

Find all local maximum and minimum points \((x, y)\) by the method of this section. $$ f(x)=x^{2}-98 x+4 $$

Step-by-Step Solution

Answer

The local minimum point is \((49, -2397)\).

Step 1: Find the derivative

To find the local maximum and minimum points, we first need to compute the first derivative of the function. Given the function \(f(x) = x^2 - 98x + 4\), the derivative \(f'(x)\) is calculated by applying basic differentiation rules. \[f'(x) = \frac{d}{dx}(x^2 - 98x + 4) = 2x - 98\]

Step 2: Solve for critical points

A function has local maxima or minima at its critical points, which occur where the derivative is zero or undefined. Set the derivative equal to zero and solve for \(x\):\[2x - 98 = 0\]\[2x = 98\]\[x = 49\] This is the critical point.

Step 3: Evaluate second derivative test

To determine if the critical point is a maximum or minimum, use the second derivative test. First, find the second derivative of the function:\[f''(x) = \frac{d}{dx}(2x - 98) = 2\]The second derivative \(f''(x) = 2\) is positive, indicating that the function is concave up at \(x = 49\), so this point is a local minimum.

Step 4: Find the y-value of the local minimum

Substitute \(x = 49\) back into the original function to find the corresponding \(y\)-value:\[f(49) = (49)^2 - 98 \times 49 + 4\]\[= 2401 - 4802 + 4\]\[= -2397\]Thus, the local minimum point is \((49, -2397)\).

Key Concepts

Local Maximum and Minimum

When analyzing a function, identifying local maximum and minimum points is crucial. These points represent the highest or lowest values of a function within a specific interval. Imagine standing on a hill; the peak is a local maximum, while any valley is a local minimum. In calculus, these points help us understand where a function reaches its extremities.
A local maximum occurs at a point where the function changes direction from increasing to decreasing, and similarly, a local minimum occurs where the function changes from decreasing to increasing.

• Local Maximum: Highest point within a given interval.
• Local Minimum: Lowest point within a given interval.

To find these points, you need to investigate the function's critical points and ascertain their nature using specific tests like the second derivative test.

Derivative

In calculus, the derivative of a function tells us how the function's value changes as its input changes. It's the fundamental tool for finding rates of change and is key in pinpointing critical points. The derivative reflects the slope of the tangent line to the function's graph at any point.
The basic idea is differentiating a function provides a new function, often denoted as \( f'(x) \), which represents the rate of change or slope.

• Derivative tells us about the slope or steepness of a function.
• It is calculated by finding the limit of the function's average rate of change over an interval as the interval's width approaches zero.
• In polynomial functions, derivation generally involves using standard differentiation rules like the power rule.

For a quadratic function, the derivative is a linear function. For our exercise, \( f'(x) = 2x - 98 \) represents the slope of \( f(x) = x^2 - 98x + 4 \) at any point \( x \).

Critical Points

Critical points are where a function's derivative is either zero or undefined. These are potential points of local maxima or minima. Think of them as places on a graph where the function 'pauses' momentarily before changing direction.
Finding critical points is a multi-step process:

• Calculate the derivative of the function.
• Solve the equation where the derivative is equal to zero.

In the exercise, solving \( 2x - 98 = 0 \) gives us the critical point \( x = 49 \). Only when a critical point is verified with additional tests, such as the second derivative test, can it be classified definitively as a maximum or minimum.

Second Derivative Test

The second derivative test is a method to determine whether a critical point is a local maximum, local minimum, or neither. It's an extension of finding critical points, involving evaluation of the second derivative at those points.
The principle is straightforward:

• If the second derivative at a critical point is positive \( f''(x) > 0 \), the function is concave up, and the critical point is a local minimum.
• If the second derivative at a critical point is negative \( f''(x) < 0 \), the function is concave down, and the critical point is a local maximum.

In the provided exercise, the second derivative is constant, \( f''(x) = 2 \), which is positive, confirming that \( x = 49 \) is a local minimum. This method is reliable as it checks not just the slope but the function's curvature at the critical points.