# function help

• Jun 25th 2006, 02:36 AM
kwtolley
algebra and trigonometry with analytic geometry problem
The table shows several value of the function f(x)=-x^3+x^2-x+2. Complete the missing values in his table, and then use these values and the intermediate value theorem to determine (an) interval(s) where the function must have a zero.

table
x -2 -1 0 1 2
f(x) 16 -4
My answer is (-infty,0) u (2,infty), but I think that I'm wrong. thanks for any help.
• Jun 25th 2006, 03:17 AM
CaptainBlack
Quote:

Originally Posted by kwtolley
The table shows several value of the function f(x)=-x^3+x^2-x+2. Complete the missing values in his table, and then use these values and the intermediate value theorem to determine (an) interval(s) where the function must have a zero.

table
x -2 -1 0 1 2
f(x) 16 -4
My answer is (-infty,0) u (2,infty), but I think that I'm wrong. thanks for any help.

Fill in the table:

$
\begin{array}{c|ccccc}
{x}& {-2}&{-1}&{0}&{1}&{2}\\
{f(x)}& {16}&{5}&{2}&{1}&{-4}
\end{array}
$

Now as $x \to -\infty,\ f(x) \to +\infty$, and $x \to +\infty,\ f(x) \to -\infty$, we have no evidence for a zero for $x<-2$, or $x>2$.

Now $f(x)$ changes sign in $(1,2)$. Hence we conclude from the intermediate value theorem (and the continuity of $f(x)$) that $f(x)$ has a root in this interval.

RonL
• Jun 25th 2006, 03:48 AM
kwtolley
table messed up on me sorry
redo of the table

x -2 -1 0 1 2
f(x) 16 _ _ _ -4