Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

find f(x+h) where f(x)=x-3 and h(x)=x+2

OpenStudy (anonymous):

Would it be f(x+x+2)?

OpenStudy (anonymous):

So then (x-3)+(x-3)+2?

OpenStudy (anonymous):

Can you still help me?

OpenStudy (anonymous):

It says to simplify the answer if it needs it...but that's it :)

OpenStudy (anonymous):

Can you do anything to reduce it, or is that it?

OpenStudy (anonymous):

So it isn't like f(x+10) or something like that, where you can still solve it further?

OpenStudy (anonymous):

it isn't f((x-3)^2)+2)

OpenStudy (anonymous):

Usually if f(x)=2x+6 (say) then f(x+h) = 2(x+h) +6 and h(x) is irrelevant, Are you doing some beginning calculus and learning about limits?

OpenStudy (anonymous):

it is the class before pre-calc...

OpenStudy (anonymous):

In the form that you asked the question . F(x+h) = x+h -3

OpenStudy (anonymous):

how?

OpenStudy (anonymous):

I replaced the "x" in the original function with "x+h".

OpenStudy (anonymous):

explain...?

OpenStudy (anonymous):

\[f(x) = x- 3\] Here replace x with x + h

OpenStudy (anonymous):

What answer were you trying to get for this question?

OpenStudy (anonymous):

\[f(x+h) = (x+h) - 3\]

OpenStudy (anonymous):

Are you sure your question is not this : Find f(x) + h(x) ??

OpenStudy (anonymous):

Or even f(h(x))?

OpenStudy (anonymous):

it is f(x+h)

OpenStudy (anonymous):

Then my answer agrees with waterineyes

OpenStudy (anonymous):

but why do I replace the x with x+h?

OpenStudy (anonymous):

See, f(x) = x - 3 (it simply means that f(x) is function of x) See : for x = 1 f(x) = 1 - 3 = -2 for x = 2, f(x) = 2 - 3 = -1 for x = 3 f(x) = 3 - 3 = 0 Here you can see f(x) depends upon x.. As x changes f(x) changes...

OpenStudy (anonymous):

And now f(x+h) means f(x+h) depends upon x + h: So to find it f(x) you have earlier, here if we add h to it then it will become function of x + h : So : replace x by x + h.. f(x) = x - 3 replace : f(x+h) = (x+h) - 3

OpenStudy (anonymous):

ok...I think that I get it...

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!