Question

Graph each function. Be sure to label three points on the graph. If \(f(x)=\left\\{\begin{array}{ll}x^{3} & \text { if }-2 \leq x<1 \\ 3 x+2 & \text { if } 1 \leq x \leq 4\end{array}\right.\) find: (a) \(f(-1)\) (b) \(f(0)\) (c) \(f(1)\) (d) \(f(3)\)

Short answer

f(-1) = -1, f(0) = 0, f(1) = 5, f(3) = 11

Step-by-step solution

Understand the Function Definition

The function is defined in two parts based on the value of x. For \(-2 \leq x < 1\), use the function \(f(x) = x^3 \). For \(1 \leq x \leq 4\), use the function \(f(x) = 3x + 2\).

Evaluate f(-1)

Since \(-2 \leq -1 < 1\), use the first part of the function: \(f(-1) = (-1)^3 = -1\). So, \(f(-1) = -1\).

Evaluate f(0)

Since \(-2 \leq 0 < 1\), use the first part of the function: \(f(0) = 0^3 = 0\). So, \(f(0) = 0\).

Evaluate f(1)

Since \(1 \leq 1 \leq 4\), use the second part of the function: \(f(1) = 3 \cdot 1 + 2 = 5\). So, \(f(1) = 5\).

Evaluate f(3)

Since \(1 \leq 3 \leq 4\), use the second part of the function: \(f(3) = 3 \cdot 3 + 2 = 11\). So, \(f(3) = 11\).

Key concepts

Function Evaluation

Piecewise functions are functions that have different expressions based on the input value's interval. In this exercise, the function changes its rule depending on the value of **x**. Let's evaluate the function step-by-step.

For **a)**, we need to find **f(-1)**. Since **−2 ≤ −1 < 1**, use the rule **f(x) = x³**. Substituting −1 for **x** gives us **f(-1) = (-1)³ = -1**. Therefore, **f(-1) = -1**.

For **b)**, we need to find **f(0)**. Similarly, since **−2 ≤ 0 < 1**, use the rule **f(x) = x³**. Substituting 0 for **x** gives us **f(0) = 0³ = 0**. Therefore, **f(0) = 0**.

For **c)**, we need to find **f(1)**. Here, **1 ≤ 1 ≤ 4**, so we use the rule **f(x) = 3x + 2**. Substituting 1 for **x** gives us **f(1) = 3*1 + 2 = 5**. Therefore, **f(1) = 5**.

For **d)**, we need to find **f(3)**. As before, since **1 ≤ 3 ≤ 4**, use the rule **f(x) = 3x + 2**. Substituting 3 for **x** gives us **f(3) = 3*3 + 2 = 11**. Therefore, **f(3) = 11**.

Graphing Functions

Graphing a piecewise function involves graphing each part on its specific interval. In this exercise, we have two rules: **x³ for −2 ≤ x < 1** and **3x + 2 for 1 ≤ x ≤ 4**.

To graph the first part, **x³**, focus on the interval **−2 ≤ x < 1**. Select a few **x** values in this range, such as −2, −1, 0. Plug these values into **x³** to get:
• **f(-2) = (-2)³ = -8**
• **f(-1) = (-1)³ = -1**
• **f(0) = 0³ = 0**
Plot these points: **(-2, -8), (-1, -1), (0, 0)**. Connect them smoothly since **x³** is a cubic function.

Next, graph **3x + 2** for **1 ≤ x ≤ 4**. Again, select a few **x** values like 1, 2, 3, and 4. Plug these values into **3x + 2** to get:
• **f(1) = 3*1 + 2 = 5**
• **f(2) = 3*2 + 2 = 8**
• **f(3) = 3*3 + 2 = 11**
• **f(4) = 3*4 + 2 = 14**
Plot those points: **(1, 5), (2, 8), (3, 11), (4, 14)**, and connect them with a straight line since **3x + 2** is a linear function.

The final graph will show a smooth cubic curve on the left of **x = 1** and a straight line starting from **(1, 5)** to **(4, 14)**.

Step-by-Step Solution

To build strong foundations, let's break down each step of the problem. This will help you follow the logical progression for solving piecewise functions.

**Step 1: Understand the Function Definition**
Recognize that the function is defined in two parts, changing at **x = 1**. Carefully observe which part to use based on the value of **x**.

**Step 2: Evaluate f(-1)**
Identify that **−2 ≤ −1 < 1** falls under the rule **f(x) = x³**. Substituting -1 gives us **f(-1) = -1**.

**Step 3: Evaluate f(0)**
Note that **0** is in the interval **−2 ≤ 0 < 1**, so use **f(x) = x³** again. Substituting **0** gives **f(0) = 0**.

**Step 4: Evaluate f(1)**
Since **1 ≤ 1 ≤ 4**, switch to using **f(x) = 3x + 2**. Substituting **1** results in **f(1) = 5**.

**Step 5: Evaluate f(3)**
For **3**, which falls in the interval **1 ≤ 3 ≤ 4**, use **f(x) = 3x + 2**. Substituting **3** delivers **f(3) = 11**.

Following these steps methodically ensures a correct evaluation of piecewise functions. Use this process for any similar problems to simplify and accurately solve them.