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By admin 数学代写 2022年2月23日

$7.9$ Exercises

Sally drives her Saturn over the 110 mile toll road in exactly $1.3$ hours. The speed limit on this toll road is 70 miles per hour and the fine for speeding is 10 dollars per mile per hour over the speed limit. How much should Sally pay?
Two cars are careening down a freeway in Utah weaving in and out of traffic. Car A passes car B and then car B passes car A as the driver makes obscene gestures. This infuriates the driver of car A who passes car B while firing his handgun at the driver of car B. Show there are at least two times when both cars have the same speed. Then show there exists at least one time when they have the same acceleration. The acceleration is the derivative of the velocity.
Show the cubic function $f(x)=5 x^{3}+7 x-18$ has only one real zero.
Suppose $f(x)=x^{7}+|x|+x-12$. How many solutions are there to the equation, $f(x)=0$ ?
Let $f(x)=|x-7|+(x-7)^{2}-2$ on the interval $[6,8]$. Then $f(6)=0=f(8)$. Does it follow from Rolle’s theorem that there exists $c \in(6,8)$ such that $f^{\prime}(c)=0$ ? Explain your answer.
Suppose $f$ and $g$ are differentiable functions defined on $\mathbb{R}$. Suppose also that it is known that $\left|f^{\prime}(x)\right|>\left|g^{\prime}(x)\right|$ for all $x$ and that $\left|f^{\prime}(t)\right|>0$ for all $t$. Show that whenever $x \neq y$, it follows $|f(x)-f(y)|>|g(x)-g(y)|$. Hint: Use the Cauchy mean value theorem, Theorem 7.8.2.
Show that, like continuous functions, functions which are derivatives have the intermediate value property. This means that if $f^{\prime}(a)<0<f^{\prime}(b)$ then there exists $x \in(a, b)$ such that $f^{\prime}(x)=0$. Hint: Argue the minimum value of $f$ occurs at an interior point of $[a, b]$.
Find an example of a function which has a derivative at every point but such that the derivative is not everywhere continuous.

CHAPTER 7. THE DERIVATIVE
144 9. Consider the function $$ f(x) \equiv\left{\begin{array}{c}1 \text { if } x \geq 0 \ -1 \text { if } x<0\end{array}\right. $$

Suppose $c \in I$, an open interval and that a function $f$, defined on $I$ has $n+1$ derivatives. Then for each $m \leq n$ the following formula holds for $x \in I .$
$$
f(x)=\sum_{k=0}^{m} f^{(k)}(c) \frac{(x-c)^{k}}{k !}+f^{(m+1)}(y) \frac{(x-c)^{m+1}}{(m+1) !}
$$
where $y$ is some point between $x$ and $c$. Fix $c, x$ in $I$. Let $K$ be a number, depending on $c, x$ such that
$$
f(x)-\left(f(c)+\sum_{k=1}^{n} \frac{f^{(k)}(c)}{k !}(x-c)^{k}+K(x-c)^{n+1}\right)=0
$$
Now the idea is to find $K$. To do this, let
$$
F(t)=f(x)-\left(f(t)+\sum_{k=1}^{n} \frac{f^{(k)}(t)}{k !}(x-t)^{k}+K(x-t)^{n+1}\right)
$$
Then $F(x)=F(c)=0$. Therefore, by Roll
e’s theorem there exists $y$ between $c$ and $x$ such that
$F^{\prime}(y)=0$. Do the differentiation and solve for $K$.
This is the main result on Taylor polynomials approximating a f
unction $f$. The term $f^{(m+1)}(y) \frac{(x-c)^{m+1}}{(m+1)
!}$ is called the Lagrange form of the remainder.
Let $f$ be a real continuous function defined on the interval $[0,1]$. Also suppose $f(0)=0$ and $f(1)=1$ and $f^{\prime}(t)$ exists for all $t \in(0,1)$. Show there exists $n$ distinct points $\left{s_{i}\right}_{i=1}^{n}$ of the interval such that
$$
\sum_{i=1}^{n} f^{\prime}\left(s_{i}\right)=n .
$$
Hint: Consider the mean value theorem applied to successive pairs in the following sum.
$$
f\left(\frac{1}{3}\right)-f(0)+f\left(\frac{2}{3}\right)-f\left(\frac{1}{3}\right)+f(1)-f\left(\frac{2}{3}\right)
$$
Now suppose $f:[0,1] \rightarrow \mathbb{R}$ is continuous and differentiable on $(0,1)$ and $f(0)=0$ while $f(1)=1$. Show there are distinet points $\left{s_{i}\right}_{i=1}^{n} \subseteq(0,1)$ such that
$$
\sum_{i=1}^{n}\left(f^{\prime}\left(s_{i}\right)\right)^{-1}=n
$$
Hint: Let $0=t_{0}<t_{1}<\cdots<t_{n}=1$ and pick $x_{i} \in f^{-1}\left(t_{i}\right)$ such that these $x_{i}$ are increasing and $x_{n}=1, x_{0}=0$. Explain why you can do this. Then argue
$$
t_{i+1}-t_{i}=f\left(x_{i+1}\right)-f\left(x_{i}\right)=f^{\prime}\left(s_{i}\right)\left(x_{i+1}-x_{i}\right)
$$
and so
$$
\frac{x_{i+1}-x_{i}}{t_{i+1}-t_{i}}=\frac{1}{f^{\prime}\left(s_{i}\right)}
$$
Now choose the $t_{i}$ to be equally spaced.
Show that $(x+1)^{3 / 2}-x^{3 / 2}>2$ for all $x \geq 2$. Explain why for $n$ a natural number larger than or equal to 1 , there exists a natural number $m$ such that $(n+1)^{3}>m^{2}>n^{3}$. Hint: Verify directly for $n=1$ and use the above inequality to take care of the case where $n \geq 2$. This shows that between the cubes of any two natural numbers there is the square of a natural number.