Algebra
posted by jessica .
Here is the problem:
Given f(x)=x^3+3x^22x+6, find f(1), f(0), and f(1)
I believe the way you solve this is replace the value of f into the equation for x right?
well that is the way I did it and I got
f(1)=6
f(0)=6
f(1)=5
My instructor said that it is wrong. Can someone help me with this??
I got
f(1)=10
f(0)=6
f(1)=8
why don't you show your steps so I can tell you where you are going wrong.
my first one is done this way
f(1) = (1)^3 + 3(1)^2  2(1) + 6
= 1 + 3 + 2 + 6
= 10
(x^3+3x^2)(2x+6)
(1+3(1)^2(2(1)+6)
(1+3)(2+6)
28
6
(x^3+3x^2)(2x+6)
(1+3(1)^2)(2(1)+6)
(1+3)(2+6)
28
6
(0^3+3(0^2)(2(0)+6)
(3)(2+6)
38
5
This is how I got those answers
why are you putting brackets around the last two terms?
they were
....  2x + 6
by placing the brackets the way you did, you are saying ...  2x  6
also the original the way you typed it was
f(x)=x^3+3x^22x+6
I did not notice the  in front of the first term, it printed so small
here is how I would do your question
f(1) = (1)^3 + 3(1)^2  2(1) + 6
= 1 + 3 + 2 + 6 = 12
f(0) = 0 + 0 + 0 + 6 = 6
f(1) = (1)^3 + 3(1)^2  2(1) + 6
= 1 + 3  2 + 6 = 6
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