For the given functions, write the rational function,
r(x) = [f(x-h)-f(x)]/h and simplify:
a. F(x) = 2x+3
b. F(x) = 1/(x+1)
c. F(x) = x^2
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Just need to know that the rational function is and maybe I can do the questions. Thanks.
the rational function is r(x)=[f(x-h)-f(x)]/h (also known as the limit definition)
so if your function is F(x)=2x+3 plug it into the rational function
[2(x-h)+3-(2x+3)]/h
=[2x+2hx-3-2x-3]/h --->then simplify
=2h/h--->cancel the h's
=2
then do the same thing for the other problems
ah sorry... I was looking at your problem and thinking of something else. Here is the correct solution for part a
so therefore...
SO if F(x)=2x+3 plug it into your rational function r(x)=[f(x-h)-f(x)]/h
=[2(x-h)+3-(2x+3)]/h
=[2x-2h+3-2x-3]/h --->then simplify (cancel the 2x's and the 3's)
=-2h/h--->cancel the h's
=-2
Sorry about that screw-up. Hope you understand!
recommended steps for factoring a polynomial
To find the rational function, r(x), for the given functions, we need to substitute f(x-h) and f(x) into the formula r(x) = [f(x-h) - f(x)]/h and simplify.
a. F(x) = 2x + 3:
Substituting this into the formula, we have:
r(x) = [f(x-h) - f(x)]/h
= [2(x-h) + 3 - (2x + 3)]/h
= (2x - 2h + 3 - 2x - 3)/h
= (2x - 2x - 2h + 3 - 3)/h
= (-2h)/h
= -2
Therefore, the rational function for F(x) = 2x + 3 simplifies to r(x) = -2.
b. F(x) = 1/(x + 1):
Substituting this into the formula, we have:
r(x) = [f(x-h) - f(x)]/h
= [1/(x-h + 1) - 1/(x + 1)]/h
At this point, we could attempt to simplify the expression, but we can also leave it as is since it cannot be simplified any further.
Therefore, the rational function for F(x) = 1/(x + 1) is r(x) = [1/(x-h + 1) - 1/(x + 1)]/h.
c. F(x) = x^2:
Substituting this into the formula, we have:
r(x) = [f(x-h) - f(x)]/h
= [(x-h)^2 - x^2]/h
= [(x^2 - 2hx + h^2) - x^2]/h
= [x^2 - 2hx + h^2 - x^2]/h
= (-2hx + h^2)/h
= -2x + h
Therefore, the rational function for F(x) = x^2 simplifies to r(x) = -2x + h.