# 建築系微積分 Exercise 3 Solution
###### tags: `Calculus` `微積分` `Arch` `建築系`

解答僅供參考，有問題請利用旁邊留言的功能、或者是和助教聯繫。

### Section 1.5
* Find the limit (if it exists)

**44.**$$\lim_{\Delta x\rightarrow0}\frac{-3(x+\Delta x)+3x}{\Delta x}$$
***Solution:***
$$\lim_{\Delta x\rightarrow0}\frac{-3(x+\Delta x)+3x}{\Delta x}=\lim_{\Delta x\rightarrow0}\frac{-3\Delta x}{\Delta x}=\lim_{\Delta x\rightarrow0}-3\ (\Delta x\neq0)=-3$$
**48.**$$\lim_{x\rightarrow3}\frac{\sqrt{x+1}-2}{x-3}$$
***Solution:***
$$\lim_{x\rightarrow3}\frac{\sqrt{x+1}-2}{x-3}
=\lim_{x\rightarrow3}\frac{\sqrt{x+1}-2}{x-3}\cdot\frac{\sqrt{x+1}+2}{\sqrt{x+1}+2}
=\lim_{x\rightarrow3}\frac{x-3}{(x-3)(\sqrt{x+1}+2)}
=\lim_{x\rightarrow3}\frac{1}{\sqrt{x+1}+2}=\frac{1}{4}$$

**56.**$$\lim_{x\rightarrow5}\frac{4}{x-5} $$
***Solution:***
We obeserve that the limit from the left $$\lim_{x\rightarrow5^-}\frac{4}{x-5}=-\infty $$ and the limit from the right $$\lim_{x\rightarrow5^+}\frac{4}{x-5}=\infty $$ are both unbounded, so the limit does not exist.

### Section 1.6
* Use a graphing utility to graph the function. Use the graph to determine any $x$-value(s) at which the function is not continuous. Explain why the function is not continuous at the $x$-value(s).

**54.**
$$f(x)=\left\{\begin{array}{ll}
3x-2, & x\leq2 \\
x+1, & x>2
\end{array}\right.$$
***Solution:***
Observe from the [graph of $f$](https://www.desmos.com/calculator/yyboxb0p3o), we find that $f$ is not continuous on $x=2$, since the limit $\lim_{x\rightarrow2}f(x)$ does not exist.
To see this, we compute both limit from the right: $$\lim_{x\rightarrow2^+}f(x)=3$$
and the limit from the left: $$\lim_{x\rightarrow2^-}f(x)=4.$$

**68.** The number of units in inventory in a small company is
$$N(t)=25\left(2\left[\frac{t+2}{2}\right]-t\right)$$
where $t$ is the time in months.
a. Use the greatest integer function $[\cdot]$ of a graphing utility to graph this function and then discuss its continuity.
b. Find the number of units in inventory during the seventh month.
c. How often must the company replenish its inventory?
***Solution:***
a. Obeserve the [graph of $N(t)$](https://www.desmos.com/calculator/3uga3foliq), we find it is not continuous at even integer $t$.
b. Let $t=7$, and we have $$N(7)=25\left(2\left[\frac{7+2}{2}\right]-7\right)=25.$$
c. From the graph of $N$, we find $N\rightarrow0$ when $t\rightarrow2^-,\ 4^-,\ 6^-,\ 8^-,\ 10^-,\ 12^-$, so the company should replenish its inventory every two months.

### Section 2.1
* Find a equation of the tangent line to the graph of $f$ at the given point. Then verify your results by using a graphing utility to graph the function ad its tangent line at the point.

**45.** $f(x)=\sqrt{x}+1;\ (4,3)$
***Solution:***
We first find the slope of the tangent line by computing the limit
$$\lim_{h\rightarrow0}\frac{f(4+h)-f(3)}{h}
=\lim_{h\rightarrow0}\frac{(\sqrt{4+h}+1)-3}{h}
=\lim_{h\rightarrow0}\frac{\sqrt{4+h}-2}{h}\cdot\frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}
=\lim_{h\rightarrow0}\frac{1}{\sqrt{4+h}+2}=\frac{1}{4}.
$$
Therefore we have the equation of tangent line to $f$ at $(4,3)$:
$$(y-3)=\frac{1}{4}(x-4)\Rightarrow
x-4y+8=0.$$
From the [graph](https://www.desmos.com/calculator/gpyapwawr3), we can see that this is exactly a tangent line to $f$ at $(4,3)$.

**48.** $f(x)=\frac{1}{x-3};\ (2,-1)$
***Solution:***
We fisrt find the slope of the tangent line by computing the limit
$$\lim_{h\rightarrow0}\frac{f(2+h)-f(2)}{h}
=\lim_{h\rightarrow0}\frac{\frac{1}{2+h-1}+1}{h}
=\lim_{h\rightarrow0}\frac{1}{h}\left(\frac{1+(h-1)}{h-1}\right)
=\lim_{h\rightarrow0}\frac{1}{h-1}=-1
$$
Therefore we have the equation of tangent line to $f$ at $(2,-1)$:
$$(y+1)=-(x-2)\Rightarrow
x+y-1=0.$$
From the [graph](https://www.desmos.com/calculator/zj9rvcl6di), we can see that this is exactly a tangent line to $f$ at $(2,-1)$.

* Find an equation(s) of the line(s) that is tangent to the graph of $f$ and parallel to the given line.

**52.**
Function: $f(x)=x^3+2$
Line: $3x-y-4=0$
***Solution:***
We find the derivative of $f$:
$$\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}
=\lim_{h\rightarrow0}\frac{(x+h)^3-x^3}{h}$$
$$=\lim_{h\rightarrow0}\frac{h[(x+h)^2+(x+h)\cdot x+x^2]}{h}
=\lim_{h\rightarrow0}(x+h)^2+(x+h)\cdot x+x^2=3x^2.
$$
Then we solve $3x^2=3$ and have $x=\pm1$. So there are two tagent lines parallel to the given line:
$3x-y=0$, and $3x-y+4=0$.

* Determine whether the statement is true or false. If it is false explain why or give an example that shows it is false.

**69.** The slope of the graph $y=x^2$ is different at every point on the graph of $f$.
***Solution:***
We find the drivative of $f(x)=x^2$:
$$\lim_{h\rightarrow0}\frac{(x+h)^2-x^2}{h}
=\lim_{h\rightarrow0}\frac{h(2x+h)}{h}
=\lim_{h\rightarrow0}2x+h=2x.
$$
Also, we find that the function $2x$ is different for different input $x$. Therefore, the statement is true.

**70.** If a function is continuous at a point, then it is differentiable at that point.
***Solution:***
The statement is false. We consider the function $f(x)=|x|$ at origin. From observing the graph of $f$, $f$ is continuous at origin. However, the limit $\lim_{h\rightarrow0}\frac{|h|} {h}$ does not exist. Since $\lim_{h\rightarrow0^+}\frac{|h|} {h}=1$ and $\lim_{h\rightarrow0^-}\frac{|h|} {h}=-1$.
### Section 2.2
**47.**
a. Find an equation of the tangent line to the graph of the function at the given point.
b. Use a graphing utility to graph the function and its tangent line at the point.
c. Use the *tangent* feature of a graphing utility to confirm your results.
$f(x)=\sqrt[3]{x}+\sqrt[5]{x}$; $(1,2)$
***Solution:***
a. We find derivative of $f$ at $(1,2)$:
$$\frac{d}{dx}f(x)=\frac{d}{dx}[\sqrt[3]{x}+\sqrt[5]{x}]=\frac{1}{3}x^{-\frac{2}{3}}+\frac{1}{5}x^{-\frac{4}{5}} $$
$$\Rightarrow f'(1)=\frac{1}{3}+\frac{1}{5}=\frac{8}{15}. $$
So the equation of the tagent line of $f$ at $(1,2)$ is
$$y-2=\frac{8}{15}(x-1)\Rightarrow y=\frac{8}{15}x+\frac{22}{15}.$$
b.c. You can check the graph of $f$ [here](https://www.desmos.com/calculator/wrktytz2xa).

**64.** Determine the point(s), if any, at which the graph of the function has a horizontal tangent line.
$y=x^3+3x^2$
***Solution:***
We denote $f(x)=x^3+3x^2$ and we find its derivative:
$$\frac{d}{dx}f(x)=\frac{d}{dx}(x^3+3x^2) =3x^2+6x.
$$
Then we solve the equation $3x^2+6x=0$ and we have $3x(x+2)=0$, so the function has horizontal tangent line at $(0,0)$ and $(-2,4)$.

**80.** A politician raises funds by selling tickets to a dinner for $500. The politician pays $150 for each dinner and has fixed costs of $7000 to rent a dining hall and wait staff. Write the profit $P$ as a function of $x$, the number of dinners sold. Show that the derivative of the profit is a constant and is equal to the increase in profit from each dinner sold.
***Solution:***
The profit $P$ as a function of $x$:
$$P=500x-150x-7000=350x-7000.$$
Then we find derivative of $P$:
$$\lim_{h\rightarrow0}\frac{350(x+h)-7000-350x+7000}{h}
=\lim_{h\rightarrow0}350=350.$$
### Section 2.4
**46.** Find the derivative of the function $f(x)=(3x^3+4x)(x-5)(x+1)$. State which differentiation rule(s) you use to find the derivative
***Solution:***
$$\begin{array}{ll}
\frac{d}{dx}f(x)=\frac{d}{dx}(3x^3+4x)(x-5)(x+1)& \\
=\left[\frac{d}{dx}(3x^3+4x)\right](x-5)(x+1)+(3x^3+4x)\left[\frac{d}{dx}(x-5)(x+1)\right] & {\sf (product\ rule)}\\
={\bf (9x^2+4)}(x-5)(x+1)+(3x^3+4x)\left[\frac{d}{dx}(x-5)(x+1)\right] & {\sf (power\ rule)}\\
=(9x^2+4)(x-5)(x+1)+(3x^3+4x){\bf \left\{\left[\frac{d}{dx}(x-5)\right](x+1)+(x-5)\frac{d}{dx}[(x+1)]\right\}} & {\sf (product\ rule)}\\
=(9x^2+4)(x-5)(x+1)+(3x^3+4x)(2x-4) & \\
=15x^4-48x^3-33x^2-32x-20 &
\end{array}$$
**50.** Find an equation of the tagent line to the graph of the function at the given point. Then use a graphing utility to graph the function and the tagent line in the same viewing window.
Function: $f(x)=\sqrt{x}(x-3)$
Point: $(9,18)$
***Solution:***
We find derivative of $f$ by product rule and power rule:
$$\frac{d}{dx}f(x)=\frac{1}{2}x^{-\frac{1}{2}}(x-3)+\sqrt{x}\Rightarrow f'(9)=1+3=4. $$
So the equation of the tangent line is:
$$y-18=4(x-9)\Rightarrow y=4x-18.$$
You can check the graph [here](https://www.desmos.com/calculator/kc0rpsqpvv).