Problem 12 - The Calculus with Analytic Geometry

Question

Find the absolute extrema of the given function on the given interval, if there are any, and find the values of \(x\) at which the absolute extrema occur. Draw a sketch of the graph of the function on the interval. $$ f(x)=x^{2}-2 x+4 ;(-\infty,+\infty) $$

Step-by-Step Solution

Answer

The absolute minimum value is 3 at x = 1. There is no absolute maximum.

Step 1: Understand the function and interval

The given function is f(x) = x^2 - 2x + 4 and the interval is (-∞, ∞).

Step 2: Find the derivative

To find the critical points, first find the derivative of the function. The derivative of f(x) = x^2 - 2x + 4 is f'(x) = 2x - 2.

Step 3: Solve the derivative for zero

Set the derivative equal to zero to find the critical points. 0 = 2x - 2 Solving for x, we get: x = 1.

Step 4: Evaluate the function at critical points

Evaluate f(x)at the critical point x = 1.f(1) = (1)^2 - 2(1) + 4 = 1 - 2 + 4 = 3.

Step 5: Determine the behavior at endpoints

Since the interval is (-∞, ∞), examine the behavior of f(x) as x approaches ±∞. In both directions, the term x^2 dominates, indicating f(x)goes to ∞.

Step 6: Identify absolute extrema

Given the parabola opens upwards and the critical point at x = 1 yields the minimum value f(1) = 3, the absolute minimum of f(x)is 3 at x = 1. There are no absolute maximum values as the function grows to ∞as x approaches ±∞.

Step 7: Sketch the graph

Draw a parabola opening upwards with vertex (1, 3).The function increases without bound as x moves towards ±∞.

Key Concepts

Critical Points

Critical points are essential in finding where a function changes its increasing or decreasing behavior. They occur where the first derivative of the function equals zero or is undefined. To find the critical points of a function, first, compute the derivative. Then, solve for the values of the variable that make this derivative zero. In our exercise, we solved the derivative of the function, found that the critical point is at \(x = 1\). This is because when \( f'(x) = 2x - 2 \) is set to zero, solving for \(x\) gives us \(x = 1\). This point is crucial because it helps determine where the function might have a maximum or minimum value.

Derivative

A derivative represents the rate at which a function is changing at any given point. It's a fundamental tool for determining the slope of a function at a certain point. In our exercise, the derivative of the function \(f(x) = x^2 - 2x + 4\) was calculated as \( f'(x) = 2x - 2\).

• This tells us how steep or flat the function is at any point \(x\).
  We set this derivative equal to zero to find the critical points, because the function changes direction where its slope is zero.
• Therefore, finding the derivative is the first step in locating the critical points.

Parabola

A parabola is a symmetric curve shaped like an arch. It's the graph of a quadratic function, such as \(f(x) = x^2 - 2x + 4\).
In our exercise, evaluating the function at the critical point \(x = 1\) and observing the behavior as \(x\) approaches ±∞ confirmed the graph's shape. This parabola opens upwards because the coefficient of \(x^2\) (which is 1) is positive.

• This upward opening indicates that there is a minimum point, but no maximum point.
• At \(x = 1\), the function is at its lowest, i.e., \(f(1) = 3\), which signifies the minimum value of the graphed parabola.
• Drawing the parabola helps visualize the function's behavior over the given interval.

Interval Notation

Interval notation is a way of representing the set of values that fall within a certain range. In our exercise, the interval is \((-\rightarrow, \rightarrow)\).

• This means all values from negative infinity to positive infinity, essentially the entire real number line.
• When dealing with functions and their extrema over such an interval, examine the function's behavior as it approaches the extremes.

For the given function, as \(x\) moves towards ±∞, the term \(x^2\) dominates, causing the function to grow without bound. As a result, the function has no maximum value since it increases indefinitely.