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MATH 180\hfill FALL 1995

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%10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#28}

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4.9, \#1.  $f(x)=1/x$, $f'(x)=-1/x^2$.  In particular,
$f(1)=1$, $f'(1)=-1$, so the local linearization
at $x=1$ is $y=1-(x-1)=2-x$.

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4.9, \#6(a).
$f(x)=(1+x)^k$, $f'(x)=k(1+x)^{k-1}$.  In particular,
$f(0)=1$, $f'(0)=k$, so the local linearization
at $x=0$ is $y=1+k(x-0)=1+kx$.

\noindent 6(b).
Part (a) for $k=1/2$ gives the linearization
$y=1+.5x$ for $\sqrt{1+x}$.
Taking $x=0.1$ then gives the estimate $1.05$ for $\sqrt{1.1}$

\noindent 6(c).  The actual value is
$\sqrt{1.1}=1.0488$ up to four decimal places.
The estimate in part (b) is larger than the actual value
since $f(x)$ is concave-down.  Indeed,
$f''(x)=-{1\over4}(x+1)^{-3/2}$ is always negative.

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4.9, \#9(a).
$1/(1+x)$ is the function of \#6 with $k=-1$.  So its
linearization at $x=0$ is $1+kx=1-x$.

\noindent 9(b).
When $x$ is near 0, $x^2$ is near 0 (in fact $x^2$ is even
nearer to 0).
Substituting $x^2$ for $x$ in $1-x\approx 1/(1+x)$
gives
$$1-x^2\approx\displaystyle {1\over 1+x^2}.$$

\noindent 9(c).
The parabola $1-x^2$ has derivative 0 at $x=0$.
Thus $(1/(1+x^2))$ has derivative 0 at $x=0$.

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4.Review, \#28.
The coefficient of $h$ in the exponent of
$\displaystyle P=30e^{-3.23\times 10^{-5}h}$ is negative.
Thus the graph of $P$ is one of exponential decay;  the graph passes
through $(0,30)$.
$$dP/dh\Big|_{h=0}=30(-3.23\times 10^{-5})e^{-3.23\times 10^{-5}h}
   \Big|_{h=0}=-9.69\times 10^{-4}$$
So the tangent line at $h=0$ is
$30-9.69\times 10^{-4}h$, the units being inches of mercury.
The rule of thumb is the linear function $P=30-0.001h$.
This is esentially the equation of the tangent line since
$9.69\times 10^{-4}=0.000969$.
The approximation is less than the actual
formula since $P$ is concave-up.

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4.Review, \#32. $f(x)=a^x$, $f'(x)=a^x\,\ln a$.
The tangent line to $a^x$ at $x=0$ is $1+(\ln a)x$
since $f(0)=1$ and $f'(0)=\ln a$.
The lines $1+(\ln a)x$ and $x+1$ are the same only when $a=e$.
If $a=e$, $x+1$ meets the graph of $a^x$ only at $(0,1)$
since $a^x$ is concave-up.
If $a\ne e$, $x+1$ meets the graph of $a^x$ in two points.
Indeed, if $1<a<e$, then
$x+1$ meets the graph in a second point to the right of $(0,1)$
since the slope of $x+1$ is greater than the slope of $1+(\ln a)x$.
If $e<a$, then $x+1$ meets the graph in a second point
to the left of $(0,1)$ since
$x+1$ has smaller slope than the slope of $1+(\ln a)x$.


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