Answered: (x-l for x < -1 f (x) = {ax +b for – 1…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section: Chapter Questions

Problem 5T

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Question

Find the value of the constant ( a, b, or c) that makes the function continuous.

(x-l
for x < -1
f (x)
= {ax +b for – 1<x <;
for x >
}
1/2

Expert Solution

Step 1

Given: $f (x) = \{\begin{cases} \frac{1}{x}, & x < - 1, x > \frac{1}{2} \\ a x + b, & - 1 \leq x \leq \frac{1}{2} \end{cases}$

if given function is continuous then

$\begin{array}{rcl} \lim_{x \to - 1^{-}} f (x) & = & \lim_{x \to - 1^{+}} f (x) . . (1) \\ \lim_{x \to {\frac{1}{2}}^{-}} f (x) & = & \lim_{x \to {\frac{1}{2}}^{+}} f (x) . . (2) \end{array}$

Step 2

so, by using equation $(1)$

$\begin{array}{rcl} \lim_{x \to - 1^{-}} f (x) & = & \lim_{x \to - 1^{+}} f (x) \\ \frac{1}{- 1} & = & a (- 1) + b \\ - 1 & = & - a + b \\ a - b & = & 1 . . (3) \end{array}$

similarly, by using equation $(2)$

$\begin{array}{rcl} \lim_{x \to {\frac{1}{2}}^{-}} f (x) & = & \lim_{x \to {\frac{1}{2}}^{+}} f (x) \\ a (\frac{1}{2}) + b & = & \frac{1}{(\frac{1}{2})} \\ \frac{a}{2} + b & = & 2 \\ \frac{a + 2 b}{2} & = & 2 \\ a + 2 b & = & 4 . . (4) \end{array}$

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(x-l for x < -1 f (x) = {ax +b for – 1 } 1/2