Find the values of a and b that make f continuous everywhere. a = II f(x) = x² - 4 X-2 ax² 4x = a + b ·bx + 3 if x < 2 if 2 ≤ x < 3 if x ≥ 3

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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Q12. Please answer all the parts to this question

The image presents a mathematical problem requiring the determination of specific values for the variables \( a \) and \( b \) to ensure that the function \( f(x) \) is continuous across its entire domain. The function \( f(x) \) is defined piecewise as follows:

\[ 
f(x) = 
\begin{cases} 
\dfrac{x^2 - 4}{x - 2} & \text{if } x < 2 \\
ax^2 - bx + 3 & \text{if } 2 \leq x < 3 \\
4x - a + b & \text{if } x \geq 3 
\end{cases}
\]

There are also blank spaces provided to fill in the values of \( a \) and \( b \):

\[ 
a = \_\_\_\_ 
\]

\[ 
b = \_\_\_\_ 
\] 

The goal is to solve for \( a \) and \( b \) such that \( f(x) \) is continuous everywhere.
Transcribed Image Text:The image presents a mathematical problem requiring the determination of specific values for the variables \( a \) and \( b \) to ensure that the function \( f(x) \) is continuous across its entire domain. The function \( f(x) \) is defined piecewise as follows: \[ f(x) = \begin{cases} \dfrac{x^2 - 4}{x - 2} & \text{if } x < 2 \\ ax^2 - bx + 3 & \text{if } 2 \leq x < 3 \\ 4x - a + b & \text{if } x \geq 3 \end{cases} \] There are also blank spaces provided to fill in the values of \( a \) and \( b \): \[ a = \_\_\_\_ \] \[ b = \_\_\_\_ \] The goal is to solve for \( a \) and \( b \) such that \( f(x) \) is continuous everywhere.
Expert Solution
Step 1

Given function:-

fx=x2-4x-2if x<2ax2-bx+3if 2x<34x-a+bif x3

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