# Homework 11

Exercise: Use the mean value theorem to establish: 1/8<\sqrt{51}-7<1/7.

Proof. Consider f(x)=\sqrt{x}. By the mean value theorem, there is x\in(49,51) such that f'(x)=f(51)-f(49)/2. Notice that f'(x)=\frac{1}{2\sqrt{x}}. Hence \sqrt{51}-7=f(51)-f(49)=1/\sqrt{x}\in(1/\sqrt{51},1/\sqrt{49})\subset(1/8,1/7).

Exercise: Prove that at most one of the following functions f and g can be a derivative of a real-valued function on \mathbb{R}.

Proof. We prove a stronger result that g is not a derivative of any real-valued function.

Suppose for sake of contradiction g(x)=G'(x). Then -g(-x)=G'(-x). But g(x)+g(-x), on one hand, is the derivative of G(x)-G(-x), and on the other hand, always equals 0 except at the point 0. This contradicts the immediate value theorem for derivates.

Exercise: Suppose f,g:[a,b]\rightarrow\mathbb{R} are differentiable on [a,b], and

f(x)g'(x)-f'(x)g(x)\neq 0 \text{ for all }x\in[a,b]

Suppose there are a\leq x_1<x_2\leq b with f(x_1)=f(x_2)=0. Then there is a point z\in(x_1,x_2) with g(z)=0.

Proof. First, it’s easy to see that g(x_1)\neq 0 \neq g(x_2). Suppose for sake of contradiction that g(z)\neq 0 for all z\in(x_1,x_2). Define h(x)=f(x)/g(x) on [x_1,x_2]. Note that h(x_1)=0=h(x_2). By the mean value theorem, there is z\in(x_1,x_2) such that h'(z)=0. However, 0=h'(z)=(f(z)g'(z)-f'(z)g(z))/g^2(z), which contradicts the condition given in the problem.

Exercise: Let f:[a,b]\rightarrow\mathbb{R} be differentiable and suppose f' is continuous on [a,b]. Prove that for \epsilon > 0, there exists \delta > 0 such that
\left|\frac{f(t)-f(x)}{t-x}-f'(x)\right|<\epsilon
for all points x,t\in[a,b] with 0<|t-x|<\delta.

Proof. Since f' is continuos on [a,b], f' is uniformly continuous on [a,b]. For any \epsilon > 0, there is \delta > 0, such that whenever|x-s|<\delta, |f'(x)-f'(s)|<\epsilon. Now, for all points x,t\in[a,b] with 0<|t-x|<\delta, by the mean value theorem, there is s between t and x, such that f'(s)=\frac{f(t)-f(x)}{t-x}. We have,

\left|\frac{f(t)-f(x)}{t-x}-f'(x)\right|=|f'(s)-f'(x)|<\epsilon ,

since |s-x|<|t-x|<\delta.

Exercise: Suppose f:\mathbb{R}\rightarrow\mathbb{R} is differentiable at a\in\mathbb{R}, and suppose two sequences \{x_n\} and \{y_n\} in \mathbb{R} that satisfy x_n\neq y_n, x_n\leq a\leq y_n for all n\in\mathbb{N}, and \lim_{n\rightarrow\infty}x_n=a=\lim_{n\rightarrow\infty}y_n. Prove that

\lim_{n\rightarrow\infty}\frac{f(y_n)-f(x_n)}{y_n-x_n}=f'(a).

Proof. By definition of the derivative, we have
f(y_n)-f(a)=[f'(a)+r(y_n)](y_n-a)
and
f(a)-f(x_n)=[f'(a)+r(x_n)](a-x_n)
where \lim_{x\rightarrow a}r(x)=0.

Add these two equations together, we have
f(y_n)-f(x_n)=f'(a)(y_n-x_n)+r(y_n)(y_a-a)+r(x_n)(a-x_n)
which implies,
\frac{f(y_n)-f(x_n)}{y_n-x_n}=f'(a)+r(y_n)\frac{y_n-a}{y_n-x_n}+r(x_n)\frac{a-x_n}{y_n-x_n}

Notice that both \frac{y_n-a}{y_n-x_n} and \frac{a-x_n}{y_n-x_n} are between 0 and 1. So if we take the limit over the previous equation, we will get the conclusion.