Problem 145 - Calculus Volume 1

Question

Find the critical points, maxima, and minima for the following piecewise functions. $$y=\left\\{\begin{array}{cc}{x^{2}+1} & {x \leq 1} \\ {x^{2}-4 x+5} & {x>1}\end{array}\right.$$

Step-by-Step Solution

Answer

The function has local minima at \( x = 0 \) and \( x = 2 \) with value 1.

Step 1: Find the derivatives of each piece

The first piece given is \( f(x) = x^2 + 1 \). The derivative of this piece is \( f'(x) = 2x \). The second piece is \( g(x) = x^2 - 4x + 5 \) and its derivative is \( g'(x) = 2x - 4 \).

Step 2: Solve for critical points

Critical points occur when the derivative is zero or undefined. For \( f'(x) = 2x \), setting the derivative equal to zero gives \( 2x = 0 \) so \( x = 0 \). For \( g'(x) = 2x - 4 \), setting it to zero gives \( 2x - 4 = 0 \), so \( x = 2 \). These are the candidates for critical points.

Step 3: Determine endpoints and domain issues

The piecewise function switches at \( x = 1 \). Check the behavior around this point. \( f(x) = x^2 + 1 \) at \( x = 1 \) gives 2, and \( g(x) = x^2 - 4x + 5 \) at \( x = 1 \) gives 2, ensuring continuity.

Step 4: Evaluate each critical point and endpoint

Evaluate \( f(x) \) at \( x = 0 \): \( f(0) = 0^2 + 1 = 1 \). Evaluate \( g(x) \) at \( x = 2 \): \( g(2) = 2^2 - 4\times2 + 5 = 1 \). At the transition point \( x = 1 \), check the continuity: both pieces equal 2.

Step 5: Determine maxima and minima

At \( x = 0 \), \( f(x) = 1 \), which is a local minimum since \( f'(x) \) changes from negative to positive. At \( x = 1 \), the value is 2 but it is not a maximum as the slopes suggest increasing trend from both sides. At \( x = 2 \), \( g(x) = 1 \), which is also a local minimum since \( g'(x) \) changes from negative to positive.

Key Concepts

Piecewise Functions

Piecewise functions are fascinating mathematical constructs. They are defined by different expressions based on the input value. Each function piece applies to a specific interval of the domain. Think of them as several functions stitched together to cover the entire number line.

For example, consider the function in the exercise:

• For inputs less than or equal to 1, use the expression: \(x^2 + 1\).
• For inputs greater than 1, switch to: \(x^2 - 4x + 5\).

A critical thing about piecewise functions is ensuring the overall function behaves nicely where the pieces switch over. In our case, at \(x = 1\), both expressions must yield the same result to maintain continuity. This is checked in the solution so that there's no abrupt jump at \(x=1\).

Derivatives

Derivatives are the tools used to understand the behavior of functions. They show how functions change and evolve as input values change. For piecewise functions, you'll need to calculate derivatives for each separate piece.

In our exercise, let's recap:

• For \(f(x) = x^2 + 1\), the derivative is \(f'(x) = 2x\).
• For \(g(x) = x^2 - 4x + 5\), the derivative is \(g'(x) = 2x - 4\).

Derivatives can indicate the slope of the function at any point. When the derivative is zero, you hit a flat spot, indicating potential critical points. This is where extrema, like max and min points, may be lurking. In our exercise, these critical points were initially found by setting each derivative to zero.

Maxima and Minima

Maxima and minima are key concepts in calculus, offering insights into the peak and trough points within a function's graph.

Once you identify critical points using derivatives, the next step is understanding whether these points are maxima, minima, or neither. This involves evaluating the behavior of the derivative around these critical points.For the function in the exercise:

• At \(x = 0\), the function value is the lowest at \(f(x) = 1\), signaling a local minimum. The derivative changes sign from negative to positive here, confirming the downturn to upturn transition.
• At \(x = 1\), both function pieces yield the value 2. However, since the slope from each side is positive, \(x = 1\) is not a maximum.
• At \(x = 2\), evaluate \(g(x) = 1\) again gives a local minimum, supported by a derivative sign change from negative to positive.

Maxima and minima are about pinpointing where functions reach their tallest or lowest spots, providing key insights for optimization problems and understanding function shapes.