12.26

1. • Let \(g : [0, a] \to \mathbb{R}\) be differentiable, \(g(0) = 0\), and \(|g'(x)| \leq M\) for \(x \in [0, a]\).
     Show that \(|g(x)| \leq Mx\) for \(x \in [0, a]\).

   Consider \(x\in\left(0,a\right)\), interval \([0, x]\), \(x > 0\).

   Using the Mean Value Theorem, \(\exists c \in (0, x) \text{ such that } \frac{g(x) - g(0)}{x - 0} = g'(c)\) \(\Rightarrow \frac{g(x)}{x} = g'(c)\)

   Since \(|g'(c)| \leq M\): \(|g(x)|=|g^{\prime}(c)|\cdot x\leq Mx\Rightarrow|g(x)|\leq Mx\)​

   • Let \(h : [0, a] \to \mathbb{R}\) be twice differentiable, \(h(0) = h'(0) = 0\), and \(|h''(x)| \leq M\) for \(x \in [0, a]\).
     Show \(|h(x)| \leq \frac{M x^2}{2}\) for \(x \in [0, a]\).

   \(h(x), \, x \in [0, x], \, x < a\)

   \(f(x) = \frac{x^2}{2}, \, \text{in } [0, x]\) satisfies the conditions of G.M.V Theorem

   \(\Rightarrow\exists c\in\left(0,x)\text{ such that }\left(f(x)-f(0)\right)h^{\prime}\left(c\right)=\left(h\left(x\right)-h\left(0\right)\right)\right.\cdot f^{\prime}\left(c\right)\)

   Then \(\frac{x^2}{2} \cdot h'(c) = h(x) \cdot c\) \(\Rightarrow|h(x)|=\left|\frac{h^{\prime}(c)}{c}\cdot\frac{x^2}{2}\right|\)

   I need an upper bound for \(h'(c)\). Consider the interval \([0, c]\)

   \(\text{MV Thm} \Rightarrow \exists d \in (0, c) \text{ such that } \frac{h'(c) - h'(0)}{c - 0} = h''(d)\) \(\Rightarrow\left|\frac{h^{\prime}(c)}{c}\right|=|h''(d)|\leq M\Rightarrow|h^{\prime}(c)|\leq M\cdot c\)

   Then \(|h(x)|=\left|\frac{h^{\prime}(c)}{c}\cdot\frac{x^2}{2}\right|\leq\frac{M\cdot c}{c}\cdot\frac{x^2}{2}=M\cdot\frac{x^2}{2}\)​

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2. Prove that \(1 - \frac{x^2}{2} < \cos(x) \quad \forall x \neq 0\) \(\Leftrightarrow 1 - \cos(x) < \frac{x^2}{2} \quad \forall x \neq 0\)

   \(f(x) = 1 - \cos(x), \quad g(x) = \frac{x^2}{2}\) satisfies conditions of G.M.V Thm

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   Let's consider \(x > 0\), and we are going to apply G.M.V.T in \([0, x]\).

   Then \(\exists c \in (0, x)\) such that \(\left(f(x)-f(0)\right)g^{\prime}\left(c)=f^{\prime}(c\right)\left(g(x)-g(0)\right)\)​

   \((1 - \cos(x) - 0) \cdot c = \left( \frac{x^2}{2} - 0 \right) \sin(c)\)​

   \(1 - \cos(x) = \frac{x^2}{2} \cdot \frac{\sin(c)}{c} < \frac{x^2}{2} \cdot 1 = \frac{x^2}{2}\)​

   Thus \(\sin(x) \leq x\)

   Exercise: Now take \(x < 0\) and consider \([x, 0]\).

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Second derivative test for local maximum and local minimum:

Suppose \(f'(c) = 0\).

(a) If \(f''(c) < 0 \implies f\) has a local maximum at \(x = c\).

(b) If \(f''(c) > 0 \implies f\) has a local minimum at \(x = c\).

Proof

(a) \(f''(c) = \lim_{x \to c} \frac{f'(x) - f'(c)}{x - c} = \lim_{x \to c} \frac{f'(x)}{x - c}\).

This implies \(f''(c) = \lim_{x \to c^+} \frac{f'(x)}{x - c} = \lim_{x \to c^-} \frac{f'(x)}{x - c}.\)

Then \(\lim_{x\to c^{+}}\frac{f^{\prime}(x)}{x-c}<0\implies f^{\prime}(x)<0\quad\forall x\in(c,c+\delta_{1})\).

And \(\lim_{x\to c^{-}}\frac{f^{\prime}(x)}{x-c}<0\implies f^{\prime}(x)>0\quad\forall x\in(c-\delta_{2},c).\)

Then by Criterion ( f′ ): \(\implies f \text{ has a local maximum at } x = c\).