Function f Property

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 2, Section 2.1

See attachment. Enjoy.

[Math Tutor]

Only (a).

(a) [tex]h(a+b)=2(a+b)=2a+2b=h(a)+h(b)[/tex] for all [tex]a,b[/tex]. It works.

(b) [tex]g(1+1)=4[/tex] but [tex]g(1)+g(1)=2[/tex]. Fails.

(c) [tex]F(a+b)=5a+5b-2[/tex] while [tex]F(a)+F(b)=5a+5b-4[/tex]. The constant gets counted twice, so it fails.

(d) [tex]G(1+1)=\tfrac{1}{2}[/tex] but [tex]G(1)+G(1)=2[/tex]. Fails (and it isn't even defined at [tex]x=0[/tex]).

The moral: the property [tex]f(a+b)=f(a)+f(b)[/tex] holds only for the pure proportional functions [tex]f(x)=mx[/tex] - a line through the origin. Any constant term, any power other than 1, breaks it.

[nycmath]

Only (a).

(a) [tex]h(a+b)=2(a+b)=2a+2b=h(a)+h(b)[/tex] for all [tex]a,b[/tex]. It works.

(b) [tex]g(1+1)=4[/tex] but [tex]g(1)+g(1)=2[/tex]. Fails.

(c) [tex]F(a+b)=5a+5b-2[/tex] while [tex]F(a)+F(b)=5a+5b-4[/tex]. The constant gets counted twice, so it fails.

(d) [tex]G(1+1)=\tfrac{1}{2}[/tex] but [tex]G(1)+G(1)=2[/tex]. Fails (and it isn't even defined at [tex]x=0[/tex]).

The moral: the property [tex]f(a+b)=f(a)+f(b)[/tex] holds only for the pure proportional functions [tex]f(x)=mx[/tex] - a line through the origin. Any constant term, any power other than 1, breaks it.

Very good. Thank you. Always good to hear from you.