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MATH 180\hfill FALL 1995

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10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#10}

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2.3, \#1.  The slope at every point of the graph is constant,
approximately $-2$, so the derivative function is the
constant function $y=-2$ on the interval $[-1,3]$.

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2.3, \#4. The given curve flattens out at $x=-2.5$ and $x=0$
so the derivative is zero at these points.
Its slope is positive on the intervals $[-4,-2.5]$ and
$[0,4]$, so the derivative is positive on these intervals with
magnitude corresponding to the steepness of the slope.
Similarly, the derivative is negative on $[-2.5,0]$ with
magnitude corresponding to the steepness of the slope.

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2.3, \#6.
The given curve flattens out at $x=-2$ and $x=1$
so the derivative is zero at these points.
Its slope is positive on the interval $[-3.7,-2]$.  As we move from
left to right in this interval the slope starts out steep,
becomes gradually less steep, and then flattens out at $x=-2$.
So the derivative on $[-3.7,-2]$ is a decreasing positive function
with a plunge to zero at $x=-2$.  The slope of the given curve is
negative or zero on $[-2,3.3]$.  As we move from
left to right in this interval, the slope starts at zero, becomes
negative but rises to zero at $x=1$ and then becomes increasingly
more negative.  So the derivative on $[-2,3.3]$ is a non-positive
function with zeros at $x=-2$ and $x=1$, decreasing rapidly
after $x=1$.

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2.3, \#8.  The curves are respectively (a) increasing and concave-up,
(b) increasing and concave-down, (c) decreasing and concave-up,
(d) decreasing and concave-down.

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2.3, \#18.  $f(x_3)$ is the greatest since it is the highest
of the given points, $f(x_4)$ is the least since it is the
lowest of the given points.  The steepest positive slope is $f'(x_5)$;
the steepest negative slope is $f'(x_3)$.

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