Sunday, May 28, 2017

College Algebra, Chapter 3, 3.6, Section 3.6, Problem 8

Evaluate f+g, fg, fg and fg of the function f(x)=9x2 and g(x)=x24 and find their domain

For f+g,

f+g=f(x)+g(x)f+g=9x2+x24

The radicand can't be a negative value. So we factor 9x2=(3x)(3+x) and x24=(x2)(x+2). Thus the domain of f(x)+g(x) is [3,2][2,3]

For fg

fg=f(x)g(x)fg=9x2x24

The radicand can't be a negative value. So we factor 9x2=(3x)(3+x) and x24=(x2)(x+2). Thus the domain of f(x)g(x) is [3,2][2,3]


For fg

fg=f(x)g(x)fg=(9x2)(x24)Substitute f(x)=9x2 and g(x)=x24fg=(9x2)(x24)Apply FOIL methodfg=9x236x4+4x2Combine like termsfg=13x2x436

The radicand can't be a negative value. So we factor 9x2=(3x)(3+x) and x24=(x2)(x+2). Thus the domain of f(x)g(x) is [3,2][2,3]


For fg

fg=f(x)g(x)fg=9x2x24Substitute f(x)=9x and g(x)=x24fg=9x2x24

The function fg can't have a denominator equal to zero and the radicand can't be a negative value. So we factor 9x2=(3x)(3+x) and x24=(x2)(x+2). Thus, the domain of fg is [3,2)(2,3]

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