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Consider the function f(x) = { 4x² =−x+5c_if x ≤ 2 - 4x if x > 2 For what value of c is the function continuous at 2?

Consider the function f(x) = { 4x² =−x+5c_if x ≤ 2 - 4x if x > 2 For what value of c is the function continuous at 2?

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Consider the function 

### Problem Statement Consider the function \[ f(x) = \begin{cases} 4x^2 - x + 5c & \text{if } x \leq 2 \\ 3cx^2 - 4x & \text{if } x > 2 \end{cases} \] For what value of \( c \) is the function continuous at \( x = 2 \)? *(Round your answer to two decimal places.)* ### Explanation In the given piecewise function, the goal is to find the value of \( c \) such that the function \( f(x) \) is continuous at \( x = 2 \). This means the limit of the function as \( x \) approaches 2 from both sides must be equal, and also equal to the function value at 2. 1. **Left-Hand Limit at \( x = 2 \):** \[ \lim_{x \to 2^-} f(x) = 4(2)^2 - 2 + 5c = 16 - 2 + 5c = 14 + 5c \] 2. **Right-Hand Limit at \( x = 2 \):** \[ \lim_{x \to 2^+} f(x) = 3c(2)^2 - 4(2) = 12c - 8 \] 3. **Equality for Continuity:** For continuity at \( x = 2 \), set the limits equal: \[ 14 + 5c = 12c - 8 \] 4. **Solve for \( c \):** \[ 14 + 5c = 12c - 8 \\ 14 + 8 = 12c - 5c \\ 22 = 7c \\ c = \frac{22}{7} \approx 3.14 \] Thus, the value of \( c \) that makes the function continuous at \( x = 2 \) is approximately \( 3.14 \).
Transcribed Image Text:### Problem Statement Consider the function \[ f(x) = \begin{cases} 4x^2 - x + 5c & \text{if } x \leq 2 \\ 3cx^2 - 4x & \text{if } x > 2 \end{cases} \] For what value of \( c \) is the function continuous at \( x = 2 \)? *(Round your answer to two decimal places.)* ### Explanation In the given piecewise function, the goal is to find the value of \( c \) such that the function \( f(x) \) is continuous at \( x = 2 \). This means the limit of the function as \( x \) approaches 2 from both sides must be equal, and also equal to the function value at 2. 1. **Left-Hand Limit at \( x = 2 \):** \[ \lim_{x \to 2^-} f(x) = 4(2)^2 - 2 + 5c = 16 - 2 + 5c = 14 + 5c \] 2. **Right-Hand Limit at \( x = 2 \):** \[ \lim_{x \to 2^+} f(x) = 3c(2)^2 - 4(2) = 12c - 8 \] 3. **Equality for Continuity:** For continuity at \( x = 2 \), set the limits equal: \[ 14 + 5c = 12c - 8 \] 4. **Solve for \( c \):** \[ 14 + 5c = 12c - 8 \\ 14 + 8 = 12c - 5c \\ 22 = 7c \\ c = \frac{22}{7} \approx 3.14 \] Thus, the value of \( c \) that makes the function continuous at \( x = 2 \) is approximately \( 3.14 \).
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given function f left parenthesis x right parenthesis equals open curly brackets table row cell 4 x squared minus x plus 5 c end cell cell i f space x less or equal than 2 end cell row cell 3 c x squared minus 4 x end cell cell i f space x greater than 2 end cell end table close

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