   Chapter 3.4, Problem 47E

Chapter
Section
Textbook Problem

# A function f is a ratio of quadratic functions and has a vertical asymptote x = 4 and just one x-intercept, x = 1 .It is known that f has a removable discontinuity at x = − 1 and lim x →   − 1 f ( x ) = 2 Evaluate(a) f ( 0 ) (b) lim x →   ∞ f ( x )

(a)

To determine

To evaluate:

The value of f(0)

Explanation

1) Concept:

From the given information, construct a function and evaluate the function at x=0

2) Calculation:

f is a rational function and has vertical asymptote at x=4

This shows (x-4) is one of the factors of denominator

x=1is x- intercept which shows (x-1) is one of the factor of numerator

Now, f has a removable discontinuity at x=-1

Therefore, x--1=x+1is a factor of both numerator and denominator

Therefore, the possible function is

fx=a(x-1)(x+1)(x-4)(x+1)

Now to find ause limx-1f(x)=2

limx-1a(x-1)(x+1)(x-4)(x+1)=2

Simplify,

limx-1a(x-1)(x-4)=2

Apply limit,

a(-1-1)(-1-4)=2

Simplify,

-2a-5=2

Solving for a

a=5

Therefore, function will be fx=5(x-1)(x+1)(x-4)(x+1)

At x=0

f0=5(0-1)(0+1)(0-4)(0+1)

=5(-1)(1)(-4)(1)

Simplify,

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