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12.19

  1. Tangent Line

    Determine the equation of the line tangent to the graph of \(y = x^3 - 3x + 2\) at the point \(x_0 = -2\).

    Solution
    Since \(f'(x) = 3x^2 - 3x + 0 = 3x^2 - 3\), then \(f'(-2) = 3(-2)^2 - 3 = 12 - 3 = 9\)
    Then \(y = 9x + b\), since \(f(-2)=0\), then \(0 = 9(-2) + b\), then \(b = 18\)
    Finally \(y = 9x + 18\)

  2. Normal Line

    Determine the equation of the normal line to \(y = x^3 - 3x + 2\) at the point \(x_0 = -2\).

    Suppose \(L^{\perp}:y=\tilde{m}x+\tilde{b}\)​

    Then we know \(L\) and \(L^\perp\) are orthogonal \(\iff m \cdot \tilde{m} = -1\)
    Then \(\tilde{m} = \frac{-1}{m} = \frac{-1}{9}\), then \(y = \frac{-1}{9}x + \tilde{b}\). Since \((-2, 0) \in L^\perp\), then \(0 = \frac{-1}{9}(-2) + \tilde{b}\)
    We have \(\tilde{b} = \frac{-2}{9}\), thus \(y = \frac{-1}{9}x - \frac{2}{9}\)

  3. One-to-One Function

    Let \(f\) be differentiable on an interval \(A\). Prove that if \(f'(x) \neq 0\) on \(A\), then \(f\) is injective

    To prove injection, we can take \(x_1, x_2 \in A\), \(x_1 \neq x_2\) (\(x_1 < x_2\)) and to prove \(f(x_1)\neq f(x_2)\)

    Suppose \(f(x_1)=f(x_2)\), Rolle Theorem \(\implies \exists c \in (x_1, x_2)\) such that \(f'(c) = 0\)

    But \(f'(x) \neq 0 \ \forall x \in A\), then \(f(x_1) \neq f(x_2)\)

    Therefore, \(f\) is injective.

    The converse one is not true, counter example: \(x^3\)

  4. Differentiable Function \(h\)

    Let \(h\) be a differentiable function defined on the interval \([0, 3]\), and assume that \(h(0) = 1\), \(h(1) = 2\), and \(h(3) = 2\).

    1. Argue that there exists a point \(d \in [0, 3]\) where \(h(d) = d\).

      \(f(x) = h(x) - x\) is a differentiable function on \([0, 3]\)

      \(f(0) = h(0) - 0 = 1\)
      ​\(f(1) = h(1) - 1 = 2 - 1 = 1 > 0\)
      ​\(f(3) = h(3) - 3 = 2 - 3 = -1 < 0\)

      Bolzano Theorem: \(\implies \exists d \in (1, 3)\) such that \(f(d) = 0\), then \(h(d) - d = 0\), then \(h(d) = d\) 2. Argue that at some point \(c\), we have \(h'(c) = \frac{1}{3}\).
      Since \(h(0) = 1\) and \(h(3) = 2\), then \(\frac{h(3) - h(0)}{3 - 0} = \frac{1}{3}\)

      By MVT Thm, \(\exists c \in [0, 3]\) such that \(h'(c) = \frac{1}{3}\). 3. Argue that \(h'(x) = \frac{1}{4}\) at some point in the domain.

      \(f(x)=h(x)-\frac14x+b\)

      \(f(0) = h(0) - 0 + b = 1 + b\)
      \(f(1) = h(1) - \frac{1}{4}(1) + b = 2 - \frac{1}{4} + b = \frac{7}{4} + b\)
      \(f(3) = h(3) - \frac{3}{4} + b = 2 - \frac{3}{4} + b = \frac{5}{4} + b\)

      Setting \(b = -\frac{5}{4}\): then \(f(0)=1-\frac54=-\frac14<0\), \(f(1)=\frac74-\frac54=\frac24=\frac12>0\), \(f(3) = \frac{5}{4} - \frac{5}{4} = 0\)

      By Bolzano Thm, \(\exists c \in (0, 1)\) such that \(f(c) = 0\)

      \(f(c) = f(3) = 0 \implies\) Rolle Thm \(\implies \exists \tilde{c} \in (c, 3)\) such that \(f'(\tilde{c}) = 0\)

      Since \(f'(x) = h'(x) - \frac{1}{4}\), then \(f'(\tilde{c}) = h'(\tilde{c}) - \frac{1}{4} = 0 \implies h'(\tilde{c}) = \frac{1}{4}\)

  5. Mean Value Theorem Applications

    Let \(I = (a, b)\) be an open interval, and let \(f\) be a function differentiable on \(I\). Use the Mean Value Theorem to prove the following statements:

    1. If \(f'(x) = 0\) for all \(x \in I\), then there is a constant \(r\) such that \(f(x) = r\) for all \(x \in I\).
    2. If \(f'(x) > 0\) for all \(x \in I\), then \(f(x)\) is strictly increasing on \(I\).
    3. If \(f'(x) < 0\) for all \(x \in I\), then \(f(x)\) is strictly decreasing on \(I\).
    4. Is the converse true in (2) and (3)?

  6. Inequalities

    Prove that:

    1. \(2\sqrt{x} > 3 - \frac{1}{x}\), when \(x > 1\).

      Proof

      Consider \(f(x) = 2\sqrt{x} + \frac{1}{x} - 3\). It is enough to show \(f(x) > 0 \ \forall x > 1\)

      Since \(f'(x) = \frac{1}{\sqrt{x}} - \frac{1}{x^2}\) and \(x > 1 \implies \frac{1}{\sqrt{x}} > \frac{1}{x^2}\)​\(\implies f'(x) > 0\)​\(\implies f(x)\) is monotonic increasing

      Since \(f(1) = 0\), then \(f(x) > 0 \ \forall x > 1\). 2. \(e^x > 1 + x\) for all \(x\).

      \(f(x)=(x^2-x-1)e^x\) \(Dom_f = \mathbb{R}\)

      \(f'(x) = (2x - 1)e^x + (x^2 - x - 1)e^x = (x^2 + x - 2)e^x = 0\)

      Then \(x^2 + x - 2 = 0\), then \(x = \frac{-1 \pm \sqrt{1 - 4(-2)}}{2} = \frac{-1 \pm \sqrt{9}}{2} = \frac{-1 \pm 3}{2}\) \(\Rightarrow x_1 = 1, x_2 = -2\)

      Thus Local maximum at \(x = -2\) and Local minimum at \(x = 1\)

      \(f(x) = (x^2 - x - 1)e^x\)

      \(\left(-2,f(-2)\right)=(-2,5e^2),\left(1,f\left(1\right))=(1,-e\right)\)

      \(f\) is increasing in \((-\infty, -2)\) and \((1, +\infty)\).

      \(f\) is decreasing in \((-2, 1)\).

      \(f(0) = -e^0 = -1\)

      If \(f(x) = (x^2 - x - 1)e^x = 0\), then \(x^2 - x - 1 = 0\), then \(x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\)

      image​

  7. Limits

    Determine the following limits:

    1. \(\lim_{x \to 0} \frac{x}{\tan x}\)
    2. \(\lim_{x \to 0} \frac{\ln x}{\cot x}\)
    3. \(\lim_{x \to 0} \left( \frac{1}{\sin^2 x} - \frac{1}{x^2} \right)\)
    4. \(\lim_{x \to 0} \frac{x - \arctan x}{x^3}\)

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