Problem 23 - The Calculus with Analytic Geometry

Question

\(f(x)= \begin{cases}(x+9)^{2}-8 & \text { if } x<-7 \\ -\sqrt{25-(x+4)^{2}} & \text { if }-7 \leq x \leq 0 \\ (x-2)^{2}-7 & \text { if } 0

Step-by-Step Solution

Answer

Function values are \( (x+9)^{2}-8 \( for \x < -7\), \(-\sqrt{25-(x+4)^{2}} \( for \-7 \leq x \leq 0\), \( (x-2)^{2}-7 \( for \x > 0\).

Step 1: Piecewise Function Definition

Identify and write down the segments of the piecewise function given in the exercise: - For \(x < -7\): \(f(x) = (x+9)^{2}-8\)- For -7 \leq x \leq 0\: \(f(x) = -\sqrt{25-(x+4)^{2}}\)- For \x > 0\: \(f(x) = (x-2)^{2}-7\)

Step 1: Evaluate for x < -7

To find the function value for \x < -7\, use the expression \(f(x) = (x+9)^2 - 8\). For example, if \x = -8\, then \(f(-8) = ((-8)+9)^2 - 8 =1^2 - 8 = -7\).

Step 2: Evaluate for -7 \leq x \leq 0

To find the function value for \ -7 \leq x \leq 0\, use the expression \(f(x) = -\sqrt{25-(x+4)^2}\). For example, if \x = -6\, then \(f(-6) = -\sqrt{25-((-6)+4)^2} = -\sqrt{25-4} = -\sqrt{21}\).

Step 3: Evaluate for x > 0

To find the function value for \x > 0\, use the expression \(f(x) = (x-2)^2-7\). For example, if \x = 1\, then \(f(1) = (1-2)^2 - 7 = 1 - 7 = -6\).

Key Concepts

Evaluating Piecewise Functions

Piecewise functions consist of several segments, each defined by a different algebraic expression. These segments apply to specific intervals of the domain. Evaluating a piecewise function involves determining which segment of the function corresponds to the given input value and then using the appropriate algebraic expression to find the output value. For example, in the piecewise function defined as: \[f(x)= \begin{cases}(x+9)^{2}-8 & \text { if } x<-7 \ -\text {sqrt}(25-(x+4)^{2}) & \text { if }-7 \text {≤x≤0} \ (x-2)^{2}-7 & \text { if } 0

Identify the appropriate segment

Substitute the specific value into the expression for that segment

Simplify the resulting expression

This ensures that we get the correct output for the function. To further clarify this process, let’s evaluate the function for different values within specified intervals. For example, if x = -8 (which is in the interval x < -7), we will use the formula \[f(x) = (x+9)^2 - 8\] to find \[f(-8) = ((-8)+9)^2 - 8 = 1^2 - 8 = -7\]. Similarly, if x=-6 (which is in the interval -7 ≤ x ≤ 0), we will use the formula \[f(x) = -\text{sqrt}(25-(x+4)^2)\] to find \[ f(-6)= -\text{sqrt}(25-4)= -\text{sqrt}(21)\].

Function Segments

A piecewise function is made up of multiple segments, each governed by its own unique algebraic expression. Each segment applies only to a certain interval of the function's domain. To understand function segments better, let's consider the given piecewise function: \[f(x)= \begin{cases}(x+9)^{2}-8 & \text { if } x<-7 \ -\text {sqrt}(25-(x+4)^{2}) & \text { if }-7 \text {≤x≤0} \ (x-2)^{2}-7 & \text { if } 0

The first segment, \[(x+9)^{2}-8\], applies when \[x < -7\]

The second segment, \[-\text{sqrt}(25-(x+4)^2)\], applies when \[-7 \text {≤} x \text {≤} 0\]

The third segment, \[(x-2)^{2}-7\], applies when \[0 < x\]

Applying the correct segment is important for evaluating piecewise functions accurately. By systematically breaking down the domain, we can better understand and solve for any given input. For instance, the segment \[(x-2)^2-7\] is only used when \[x>0\]. This precise allocation of segments ensures proper evaluation of piecewise functions.

Algebraic Expressions

Algebraic expressions are critical when working with piecewise functions, as each segment of the function is defined by a unique expression. These expressions can be linear, quadratic, or involve other forms such as square roots. Understanding how to manipulate and evaluate these expressions is key. For example, in the given piecewise function: \[f(x)= \begin{cases}(x+9)^{2}-8 & \text { if } x<-7 \ -\text {sqrt}(25-(x+4)^{2}) & \text { if }-7 \text {≤x≤0} \ (x-2)^{2}-7 & \text { if } 0

A quadratic expression \[(x+9)^2 - 8\] for \[x < -7\]

A square root expression \[-\text {sqrt}(25-(x+4)^2)\] for \[-7 \text {≤} x \text {≤} 0\]

Another quadratic expression \[(x-2)^2 - 7\] for \[x > 0\]

Knowing how to tackle these different algebraic forms helps simplify the process of evaluating piecewise functions. When dealing with quadratic expressions, be aware of how to expand and square binomials. When working with square roots, remember to simplify inside the root before applying the root. By handling each kind of algebraic expression properly, piecewise functions become more manageable and easier to solve.