Question

Suppose that the functions f and g are defined for all real numbers x as follows.

Answer 1

Answer:

$(f*g)(x)=2x^{2}-4x-30$

$(f-g)(x)=-(x+11)$

Step-by-step explanation:

the expression (f*g)(x) can be expanded as follows

$(f*g)(x)=f(x)*g(x)$

since $f(x)=x-5,g(x)=2x+6$

Hence

$(f*g)(x)=f(x)*g(x)\\(f*g)(x)=(x-5)*(2x+6)\\(f*g)(x)=2x^{2}+6x-10x-30\\(f*g)(x)=2x^{2}-4x-30$

also the expansion for (f-g)(x) is express as

$(f-g)(x)=f(x)-g(x)$

also since

$f(x)=x-5,g(x)=2x+6$ then when we substitute values, we have

$(f-g)(x)=(x-5)-(2x+6)\\(f-g)(x)=x-5-2x-6\\(f-g)(x)=-x-11\\(f-g)(x)=-(x+11)$

For this type of operation kindly note the below expansions

$(f*g)(x)=f(x)*g(x)$

$(f-g)(x)=f(x)-g(x)$

$(f+g)(x)=f(x)+g(x)$

$(f/g)(x)=f(x)/g(x)$