Ask your own question, for FREE!
Mathematics 10 Online
OpenStudy (anonymous):

Find (f+g),(f-g)(x), (f*g)(x), and (f/g)(x) and g(x) . f(x)=2x+1 g(x)=x-3

OpenStudy (turingtest):

do you want the process or just answers?

OpenStudy (anonymous):

process please (:

OpenStudy (turingtest):

Sure, it's all just about substituting and simplifying. Here's the first one for free just as an example: \[f(x)=2x+1\]\[g(x)=x-3\]\[(f+g)(x)=f(x)+g(x)=(2x+1)+(x-3)=3x-2\]all we did is just perform the operation (in this case addition) on each function and treated them like regular polynomials. So what do you think we will do for\[ (f-g)(x)\]????

OpenStudy (anonymous):

the same thing?

OpenStudy (anonymous):

but subtracting

OpenStudy (turingtest):

exactly, can you try it and show your work? just so I know what you need help with.

OpenStudy (anonymous):

(x-3)-(2x+1)

OpenStudy (anonymous):

= 2x^2-7x-3?

OpenStudy (turingtest):

for the first one you had it set up right for g-f, but you want f-g. You also didn't simplify, please do that over. The next one is correct. :)

OpenStudy (anonymous):

(2x+1)-(x-3)

OpenStudy (turingtest):

simplify...

OpenStudy (anonymous):

how?

OpenStudy (turingtest):

(2x+1)-(x-3)=2x+1-x+3=x+4 do you not follow that? you seemed to have no trouble multiplying the two expressions, but you can't subtract them? I think you are over-thinking this problem :)

OpenStudy (turingtest):

so what about the last one? f/g can it be simplified?

OpenStudy (anonymous):

i got that one, its 3x/-2

OpenStudy (turingtest):

no, not quite...\[{f(x)\over g(x)}={2x+1\over x-3}\]it is a direct substitution and cannot be simplified. I think you need to work on combining like terms and simplification, you seemed to have missed some important things. I'm sure you'll get it of you keep trying. Good luck!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!