• Title: Higher Derivatives

  • Series: Real Analysis

  • Chapter: Differentiable Functions

  • YouTube-Title: Real Analysis 44 | Higher Derivatives

  • Bright video: https://youtu.be/vyZ5ESoqsxw

  • Dark video: https://youtu.be/jJxvZXF8eZk

  • Quiz: Test your knowledge

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  • Subtitle on GitHub: ra44_sub_eng.srt missing

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  • Quiz Content

    Q1: Let $f \colon \mathbb{R} \rightarrow \mathbb{R}$ be a function. What is the difference between ‘differentiable’ and ‘continuously differentiable’?

    A1: There is no difference.

    A2: A differentiable function is always a continuous differentiable function but not vice versa

    A3: A continuous differentiable function is a differentiable function where $f^\prime$ is also continuous.

    A4: A continuous differentiable function is a differentiable function where $f^\prime$ is also differentiable.

    Q2: Let $f \colon \mathbb{R} \rightarrow \mathbb{R}$ be a two-times differentiable function. Is $f$ continuously differentiable?

    A1: Yes!

    A2: No!

    A3: One cannot say it.

    Q3: Let $f \colon \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x) = x^5$. Which claim is not correct?

    A1: $f$ is two-times differentiable.

    A2: $f$ is 5-times differentiable.

    A3: $f$ is 6-times differentiable.

    A4: $f$ is $\infty$-times differentiable.

    A5: $f$ has a local maximum at $x_0 = 0$.

    Q4: Let $f \colon \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x) = (x-1)^2$. Which claim is correct?

    A1: $f^{\prime \prime}(0) > 0$ implies there is a local minimum at $x_0 = 0$.

    A2: $f^{\prime \prime}(1) > 0$ implies there is a local maximum at $x_0 = 1$.

    A3: $f^\prime(1) = 0$ and $f^{\prime \prime}(1) > 0$ imply there is a local minimum at $x_0 = 1$.

    A4: $f^\prime(1) = 0$ and $f^{\prime \prime}(1) < 0$ imply there is a local minimum at $x_0 = 1$.

  • Last update: 2024-10

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