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Mathematics 16 Online
OpenStudy (anonymous):

Given the functions f(x) = 3x + 2 −4/x and g(x) = 2 − 3x, find the indicated values. a) f(a + b) b) g(a) − g(a − b) + f(a)

OpenStudy (anonymous):

replace x in the equation above with (a+b)

OpenStudy (anonymous):

f(a+b)=3(a+b)+2 -4/(a+b)

OpenStudy (xapproachesinfinity):

wherever you x replace by a+b \[f(x)=3x+2-\frac{4}{x}\] \[f(\color {red}{a+b})=3(\color{red}{a+b})+2-\frac{4}{\color{red}{a+b}}\]

OpenStudy (xapproachesinfinity):

if for example f(x)=x^-2 they ask f(c) \[f(\color{blue}{c})=\color{blue}{c}^2-2\] just a matter of replacing the dummy variable

OpenStudy (anonymous):

whats a+b tho

OpenStudy (anonymous):

what do i do when i set it up like that

OpenStudy (xapproachesinfinity):

that's what they asked you the evaluate a+b standards as some variable

OpenStudy (xapproachesinfinity):

hmm you can simply more if they ask you just depends on the question

OpenStudy (xapproachesinfinity):

in this case i would just leave it that way

OpenStudy (anonymous):

what about the second one?

OpenStudy (xapproachesinfinity):

second you gonna do each one separately g(a) g(a-b) f(a) after that the results do g(a)-g(a-b)+f(a)

OpenStudy (xapproachesinfinity):

let me see if you can do g(a)

OpenStudy (anonymous):

i thought were supposed to use the equation they gave us tho

OpenStudy (anonymous):

i dont know

OpenStudy (xapproachesinfinity):

what equation?

OpenStudy (anonymous):

the second part

OpenStudy (xapproachesinfinity):

you have g(x) right find g(a) by replacing x with a

OpenStudy (xapproachesinfinity):

i broke it down to steps as you can see first find g(a) then g(a-b) then f(a) all separately

OpenStudy (xapproachesinfinity):

try to do g(a) first let me see you do it

OpenStudy (anonymous):

so 2-3(a)?

OpenStudy (xapproachesinfinity):

yes g(a)=2-3a what about g(a-b) now

OpenStudy (xapproachesinfinity):

use the same function g(x)=2-3x to find that

OpenStudy (anonymous):

2-3(a-b)?

OpenStudy (xapproachesinfinity):

yes good

OpenStudy (xapproachesinfinity):

now f(a)

OpenStudy (anonymous):

2-3(a)-2-3(a-b)+3(a)+2-4/(a)?

OpenStudy (xapproachesinfinity):

that the next step let's not skip do f(a) first

OpenStudy (xapproachesinfinity):

i mean you have the idea but that's wrong because of the sign

OpenStudy (anonymous):

3(a)+2-4/(a)

OpenStudy (xapproachesinfinity):

yes now g(a)-g(a-b)-f(a)=2-3a-(2-3(a-b))+3a+2-4/a

OpenStudy (xapproachesinfinity):

the difference btw what i just did and what you did is i put those parenthesise before 2-3(a-b) because of the sign that came before it

OpenStudy (xapproachesinfinity):

if ignored that it would be wrong because the sign has to be distribute to 2-3(a-b)

OpenStudy (xapproachesinfinity):

so no what we do it open up the parenthesis 2-3a-2+3(a-b)+3a+2-4/a

OpenStudy (xapproachesinfinity):

we simplify more =2-3a-2+3a-3b+3a+2-4/a =....?

OpenStudy (xapproachesinfinity):

finish it up

OpenStudy (anonymous):

oh okay, thanks so much!!

OpenStudy (xapproachesinfinity):

welcome!

OpenStudy (anonymous):

so is the final answer 3a-3b+2-4/a

OpenStudy (xapproachesinfinity):

yeah seems good!

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