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MATH 180\hfill FALL 1995

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10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#14}

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2.7, \#2, 3.  The graphs of $(\sin 2x)/x$ and $(\cos x-1)/x$
on a graphing calculator show that
$$\lim_{\theta\to 0} {\sin 2\theta\over\theta}=2,\quad
  \lim_{\theta\to 0} {\cos \theta-1\over\theta}=0.$$

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2.8, \#1.  $f(x)=(x+|x|)^2+1$ is just the function
$$f(x)=\cases
    {1 &if $x<0$\cr
     4x^2+1 &if $x\ge0$\cr}$$
since $x+|x|=x-x=0$ when $x<0$ and $x+|x|=2x$ when $x\ge0$.
Its graph is the join of
the half-line $y=1$ for $x<0$ and the right half of the
parabola $y=4x^2+1$ for $x\ge0$.  These two halves meet at $(0,1)$,
so it is possible $f'(x)$ might not exist is $(0,1)$.
In a standard window $-10\le x\le10$, $-10\le y\le10$,
the graph of $f(x)$ might have a ``corner'' at $(0,1)$,
but with sufficient magnification, say
$-0.1\le x\le 0.1$, one ``sees'' there is no corner at $(0,1)$.
One can also see this directly from the definition of $f'(0)$:
$$\eqalign{
   {f(h)-f(0)\over h}
     &=\cases
        {\displaystyle {1-1\over h} &if $h<0$\cr\cr
         \displaystyle{(4h^2+1)-1\over h} &if $h>0$}\cr
     &=\cases
        {0   &if $h<0$\cr
         4h  &if $h>0$\cr}\cr}$$
As $h\rightarrow0$, the difference quotient goes to 0.
In other words, $f'(0)=0$ and $f'(0)$ exists.

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2.8, \#5.  This problem is like \#1 with minor changes.
The graph is differentiable
everywhere except possibly at $(0,0)$.  But the arguments used in \#1
apply and give $f'(0)=0$.  Example 4 on page 115 shows that the
function $x^2$ has derivative $2x$.  So the function $f(x)$
given here has derivative
$$f'(x)=\cases
    {0 &if $x<0$\cr
     2x &if $x\ge0$\cr}$$
The graph of $f'(x)$ is the join of two half-lines with a
corner at $(0,0)$.  So $f'(0)$ does not exist and
$$f''(x)=\cases
    {0 &if $x<0$\cr
     2 &if $x>0$\cr}$$



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