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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

Find the values of a and h that make f continuous everywhere.

f ( x ) = { x 2 4 x 2 if x < 2 a x 2 b x + 3 if x x < 3 2 x a + b if x 2

To determine

To find: The value of a and b which satisfies the function f(x) to be continuous everywhere.

Explanation

Given:

The function is f(x)={x24x2ifx<2ax2bx+3if2x<32xa+bifx3

Theorem used:

The functions such as “Polynomials, rational functions, root functions, trigonometric functions, inverse trigonometric functions, exponential functions and logarithmic functions” are continuous at every number in their domains.

Calculation:

The given function is polynomial and rational function. By theorem these functions are continuous on its domains. Therefore, x24x2 is continuous on (,2), ax2bx+3 is continuous on (2,3), 2xa+b is continuous on (3,).

The left hand limit of the function f(x)=x24x2 as x approaches 2 is computed as follows.

limx2f(x)=limx2(x24x2)=limx2(x222x2)=limx2((x2)(x+2)x2)=limx2(x+2)

Simplify further,

limx2f(x)=limx2x+2=2+2=4

Thus, limx2f(x)=4 (1)

The right hand limit of the function f(x)=ax2bx+3 as x approaches 2 is computed as follows.

limx2+f(x)=limx2+(ax2bx+3)=limx2+ax2limx2+bx+3=a(2)2b(2)+3=4a2b+3

Thus, limx2+f(x)=4a2b+3 (2)

Assume that the function f(x) to be continuous on the interval (,)

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