108學年度｜微積分模組班｜Class 01｜Homework 4.

Homework 4.

• Date: 108/10/14 (Week 06)
• Place: 新生教學館 301

Problem 1.

• Find the value of $a$ that makes the following function differentiable everywhere. $$f(x) = \begin{cases} ax &,\text{ if } x<0\\ x^2-3x &, \text{ if } x \geqslant 0\end{cases}$$

Solution.

• Obviously, $f(x)$ is differentiable anywhere except $x = 0$. Note that we have $$\lim_{x \to 0^-}\dfrac{f(x)-f(0)}{x-0} = a$$, and, $$\lim_{x \to 0^+}\dfrac{f(x)-f(0)}{x-0} = -3$$ In order to make $f(x)$ differentiable at $x = 0$, we have $a = -3$.

Problem 2.

• Let $f(x)$ be a strictly increasing, continuous function on $(a,b)$. Then there exists an inverse function $g(y)$ defined on $(f(a), f(b))$. Suppose furthermore that $f(x)$ is differentiable on $(a,b)$. Prove that $g(y)$ is differentiable on $(f(a), f(b))$ and $g'(y_0) = \dfrac{1}{f'(x_0)}$ if $y_0 = f(x_0)$.

Solution.

• We assume the following fact:
1. Since $f \colon (a, b) \to (f(a), f(b))$ is strictly increasing, there is an inverse function $g \colon (f(a), f(b)) \to (a, b)$ such that $g(f(x)) = x$ for all $x \in (a,b)$ and $f(g(y)) = y$ for all $y \in (f(a), f(b))$.
2. The inverse function $g(y)$ is continuous on $(f(a), f(b))$.
Fix $y_0 \in (f(a), f(b))$ with $f(x_0) = y_0$. Note that $f(x)$ is differentiable at $x = x_0$, that is, $$\lim_{x \to x_0}\dfrac{f(x)-f(x_0)}{x-x_0} = f'(x_0).$$ Since $f(x)$ is strictly increasing, so $f'(x_0) \neq 0$ and $f(x)-f(x_0) \neq 0$ whenever $x \neq x_0$, hence we have: $$\lim_{x \to x_0} \dfrac{x-x_0}{f(x)-f(x_0)} = \lim_{x \to x_0} \dfrac{1}{\dfrac{f(x)-f(x_0)}{x-x_0}} = \dfrac{1}{f'(x_0)}$$ That is, for any $\epsilon>0$, there is $\delta >0$ such that: $$\left|\dfrac{x-x_0}{f(x)-f(x_0)} - \dfrac{1}{f'(x_0)}\right|<\epsilon$ when $|x-x_0|<\delta.$$ Since $g(y)$ is continuous, there exist $\gamma >0$ such that $|x-x_0| = |g(y) - g(y_0)|<\delta$ when $|y-y_0|<\gamma$. In conclusion, when $|y-y_0|<\gamma$, we have: $$\left|\dfrac{g(y)-g(y_0)}{y-y_0} - \dfrac{1}{f'(x_0)}\right| = \left|\dfrac{x-x_0}{f(x)-f(x_0)} - \dfrac{1}{f'(x_0)}\right| < \epsilon$$ That is precisely $\displaystyle \lim_{y \to y_0}\dfrac{g(y)-g(y_0)}{y-y_0} = \dfrac{1}{f'(x_0)}$, so $g(y)$ is differentiable at $y = y_0$ and hence on $(f(a),f(b))$. 

Problem 3.

• Show that the equation $x^3 − 15x + 1 = 0$ has three solutions in the interval $[−4, 4]$.

Solution.

• Let $f(x) = x^3 − 15x + 1$, note that:
• $f(-4) = -3$, $f(-3) = 19$, by Intermediate value theorem, there is $x_1 \in (-4,-3)$ such that $f(x_1)=0$.
• $f(-3) = 19$, $f(3) = -17$, by Intermediate value theorem, there is $x_2 \in (-3,3)$ such that $f(x_2)=0$.
• $f(3) = -17$, $f(4) = 5$, by Intermediate value theorem, there is $x_3 \in (3,4)$ such that $f(x_3)=0$. 

We have exactly 3 solution in $[-4,4]$. (Polynomial of degree 3 has at most 3 roots)