Problem 50 - Thomas Calculus

Question

In Exercises \(45-52,\) find the derivative at each critical point and determine the local extreme values. $$ y=\left\\{\begin{array}{ll}{3-x,} & {x<0} \\ {3+2 x-x^{2},} & {x \geq 0}\end{array}\right. $$

Step-by-Step Solution

Answer

Local maximum at \(x = 1\), value is 4.

Step 1: Find Critical Points

First, determine where the derivative is either 0 or undefined for each piece of the piecewise function. For the first piece, \(y = 3-x\) for \(x < 0\), it is a straight line with a constant slope. The derivative is \(-1\), so there are no critical points for \(x < 0\).For the second piece, \(y = 3 + 2x - x^2\) for \(x \geq 0\), it is a quadratic function. Take the derivative to find critical points:\[\frac{dy}{dx} = 2 - 2x\]Set \(\frac{dy}{dx} = 0\) to find critical points:\[2 - 2x = 0 \implies x = 1.\]Thus, \(x = 1\) is a critical point for \(x \geq 0\).

Step 2: Evaluate the Derivative at Critical Points

Evaluate the derivative at the critical point found in Step 1. For the point \(x = 1\), substitute back into the derivative:\[\frac{dy}{dx}\bigg|_{x=1} = 2 - 2(1) = 0.\]Since the derivative is 0 at this point, it's confirmed as a critical point.

Step 3: Second Derivative Test for Local Extrema

Use the second derivative test to determine if there is a local extremum at the critical point. The second derivative of the function \( y = 3 + 2x - x^2\) is:\[\frac{d^2y}{dx^2} = -2.\]Since \(-2\) is negative, the function is concave down, implying there is a local maximum at \(x = 1\).

Step 4: Evaluate the Function at Critical Points

To find the local maximum value, evaluate the function \(y = 3 + 2x - x^2\) at \(x = 1\):\[y(1) = 3 + 2(1) - (1)^2 = 3 + 2 - 1 = 4.\]Therefore, the local maximum value is 4 at \(x = 1\).

Step 5: Check Endpoints for Local Extrema in the Interval

The endpoint for the function \(x \geq 0\) is \(x = 0\). Evaluate the piecewise function at \(x = 0\) to compare it with the critical point's value:\[y(0) = 3 + 2(0) - 0^2 = 3.\]Since \(4 > 3\), \(x = 1\) remains the local maximum, and \(x = 0\) is not a local extremum.

Key Concepts

Piecewise Functions

In mathematics, a piecewise function is defined by multiple sub-functions, each of which applies to a certain interval in its domain. Understanding these functions can initially seem daunting, but breaking them down helps.
In our exercise, the function is given in two parts, catering to different ranges of the input variable, \(x\).

• For \(x < 0\), the function is linear: \(y = 3-x\).
• For \(x \geq 0\), the function is quadratic: \(y = 3 + 2x - x^2\).

This setup shows how piecewise functions transition from one rule to the next, ensuring the whole function adapts to different values of \(x\). The idea is to manage complexities of different functions over parts of their domain, offering flexibility in mathematical modeling.

Derivative Test

Finding critical points of a function involves determining points where the derivative is zero or undefined. This process is crucial for identifying where potential maxima or minima could occur.
In the specific function analyzed, the derivative helps identify if the curve's slope changes direction, indicating possible critical points. Here's what we did:

• For the segment \(3-x\), the derivative is \(-1\), indicating a constant slope with no critical points.
• For \(3+2x-x^2\), the derivative becomes \(2 - 2x\). Setting this derivative to zero finds our critical point at \(x = 1\).

Critical points are where the curve flattens, signaling potential local minimum or maximum aspects of the graph depending on the function and context.

Local Maximum

A local maximum is a point where the function's value is higher than the values of the function in the immediate vicinity. After locating critical points using the derivative, we often employ the second derivative test to confirm the presence of a local maximum.
For the quadratic piece of the piecewise function, we followed these steps:

• At \(x = 1\), the second derivative, \(-2\), indicates the function is concave downwards (a "frown" shape), suggesting a local maximum.
• Evaluating the function at this point, we find \(y(1) = 4\), confirming the local maximum because nearby values were smaller.

The concept of a local maximum is crucial in analyzing real-world problems, enabling identification of optimal values within restricted intervals.

Quadratic Functions

Quadratic functions are polynomial functions of degree two, commonly expressed as \(ax^2 + bx + c\). They graph into a parabola that either opens upwards or downwards. Different characteristics can be derived from these functions, such as their vertex, axis of symmetry, and roots.
In our piecewise function:

• The quadratic portion, \(y=3+2x-x^2\), results in a parabola opening down because of the negative \(x^2\) term.
• Using derivatives, we identify and analyze the vertex of this parabola, which in this case, is its highest point due to the downward opening shape.

The quadratic nature adds depth to analyzing behaviour, steering us to compute precisely attributes like vertex, crucial in determining maxima or associated troughs in contextual scenarios.