Posted by jessica on Friday, May 11, 2007 at 11:54pm.
Here is the problem:
Given f(x)=-x^3+3x^2-2x+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)
2-8
-6
(-x^3+3x^2)-(2x+6)
(-1+3(1)^2)-(2(1)+6)
(-1+3)-(2+6)
2-8
-6
(-0^3+3(0^2)-(2(0)+6)
(3)-(2+6)
3-8
-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^2-2x+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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