Suppose that the functions r and a are defined for all real numbers x as follows. r(x)=2x-1 S(x)=5x write the expressions for (r-s)(x)and(r•s)(x)and evaluate(r+s)(-2).

Answer :

[tex]\boxed{(r-s)(x)=-3x-1} \\ \\ \boxed{(r\cdot s)(x)=10x^2-5x} \\ \\ \boxed{(r+s)(-2)=-15}[/tex]

Explanation:

In this exercise, we have the following functions:

[tex]r(x)=2x-1 \\ \\ s(x)=5x[/tex]

And they are defined for all real numbers x. So we have to write the following expressions:

First expression:

[tex](r-s)(x)[/tex]

That is, we subtract s(x) from r(x):

[tex](r-s)(x)=2x-1-5x \\ \\ Combine \ like \ terms: \\ \\ (r-s)(x)=(2x-5x)-1 \\ \\ \boxed{(r-s)(x)=-3x-1}[/tex]

Second expression:

[tex](r\cdot s)(x)[/tex]

That is, we get the product of s(x) and r(x):

[tex](r\cdot s)(x)=(2x-1)(5x) \\ \\ By \ distributive \ property: \\ \\ (r\cdot s)(x)=(2x)(5x)-(1)(5x) \\ \\ \boxed{(r\cdot s)(x)=10x^2-5x}[/tex]

Third expression:

Here we need to evaluate:

[tex](r+s)(-2)[/tex]

First of all, we find the sum of functions r(x) and s(x):

[tex](r+s)(x)=2x-1+5x \\ \\ Combine \ like \ terms: \\ \\ (r+s)(x)=(2x+5x)-1 \\ \\ (r+s)(x)=7x-1[/tex]

Finally, substituting x = -2:

[tex](r+s)(-2)=7(-2)-1 \\ \\ (r+s)(-2)=-14-1 \\ \\ \boxed{(r+s)(-2)=-15}[/tex]

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