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December 20, 2014

December 20, 2014

Posted by **jessica** on Friday, May 11, 2007 at 11:54pm.

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

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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