Derivative tests

Let f(x)=x3+x2+xf(x) = -x^3+x^2+x and use the first derivative test to find and classify the extreme values.

Since f is a negative cubic and we’re not given an interval, f has no absolute max or min. Here, f(x)=3x2+2x+1f’(x)=-3x^2+2x+1, which exists everywhere. Let’s set it equal to zero and solve for x:
0=3x2+2x+1=(3x+1)(x1)x=1,1/3.0 = -3x^2+2x+1 = -(3x+1)(x-1) \rightarrow x=1,-1/3.
This gives us the intervals (,1/3), (1/3,1), (1,)(-\infty, -1/3), \ (-1/3, 1), \ (1, \infty). We’ll set up a table to find the behavior of f on these intervals:
IntervalSign of f Behavior of f(,1/3)incr(1/3,1)+decr(1,)incr\begin{array}{c|c|c} \text{Interval} & \text{Sign of } f’ & \text{ Behavior of }f \\ \hline (-\infty, -1/3)& - & \text{incr} \\ (-1/3, 1)& + & \text{decr}\\ (1, \infty)& - & \text{incr} \end{array}

So f has a local min at x=1/3x=-1/3 and a local max at x=1x=1.
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