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MATH 180\hfill FALL 1995

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10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#9}

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2.2, \#6.
$f(4)>f(3)$ since $f$ in increasing.
$$f(2)-f(1)={f(2)-f(1)\over 2-1},\quad
f(3)-f(2)={f(3)-f(2)\over 3-2}$$
are slopes of lines, namely the line connecting
$(1,(f(1))$ and  $(2,f(2))$ and the line connecting
$(2,(f(2))$ and  $(3,f(3))$.
Since $f$ is concave-down, $f(2)-f(1)>f(3)-f(2)$.
In the same way we see that
$${f(2)-f(1)\over 2-1}>{f(3)-f(1)\over 3-1}$$
since the quotients are slopes of lines, namely the line conecting
$(2,f(2))$ and $(1,f(1))$ and the line connecting
$(3,f(3))$ and $(1,f(1))$.

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2.2, \#7.  $f(4)/4$ is the slope of the line connecting $(0,0)$ and
$(4,f(4))$.  Similarly, $f(3)/3$ is the slope of the line connecting
$(0,0)$ and $(3,f(3))$.  Again the concavity of $f$ shows that
$f(3)/3>f(4)/4$.

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2.2, \#10.  For $f(x)=x^3+5$
$${f(1+h)-f(1)\over h}={((1+h)^3+5)-(1+5)\over h}=
   {3h+3h^2+h^3\over h}\rightarrow 3$$
as $h\rightarrow0$.

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2.2, \#12.  For $g(t)=3t^2+5t$
$${g(-1+h)-g(-1)\over h}={(3(-1+h)^2+5(-1+h))-(3-5)\over h}=
   {-h+3h^2\over h}\rightarrow -1$$
as $h\rightarrow0$.

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Comments:
In problems \#6, 7 the quantities
in question are related to slopes of lines.  In \#6 the lines
are secant lines through two given points on the graph;  in \#7 the
lines are lines connecting the origin to given points on the graph.

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