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Use derivatives to sketch the graph of a function

Let

    \[ f(x) = \frac{1}{6}x^2  + \frac{1}{12} \cos (2x). \]

  1. Find all points such that f'(x) = 0;
  2. Determine the intervals on which f is monotonic by examining the sign of f';
  3. Determine the intervals on which f' is monotonic by examining the sign of f'';
  4. Sketch the graph of f.

  1. We take the derivative,

        \[ f'(x) = \frac{1}{3}x - \frac{1}{6} \sin (2x). \]

    Thus,

        \[ f'(x) = 0 \quad \implies \quad \frac{1}{3} x - \frac{1}{6} \sin (2x) = 0 \quad \implies \quad x = 0. \]

  2. f is increasing if x > 0 and decreasing if x < 0.
  3. Taking the second derivative,

        \[ f''(x) = \frac{1}{3} - \frac{1}{2} \cos (2x). \]

    Thus, f' is increasing for all x.

  4. We sketch the curve,

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Use derivatives to sketch the graph of a function

Let

    \[ f(x) = x + \cos x. \]

  1. Find all points such that f'(x) = 0;
  2. Determine the intervals on which f is monotonic by examining the sign of f';
  3. Determine the intervals on which f' is monotonic by examining the sign of f'';
  4. Sketch the graph of f.

  1. We take the derivative,

        \[ f'(x) = 1 - \sin x. \]

    Thus,

        \[ f'(x) = 0 \quad \implies \quad 1 - \sin x = 0 \quad \implies \quad x = \left(2n + \frac{1}{2}\right) \pi. \]

  2. f is increasing for all x since 1 - \sin x \geq 0 for all x.
  3. Taking the second derivative,

        \[ f''(x) = - \cos x. \]

    Thus, f' is increasing if \left( 2n + \frac{1}{2} \right) \pi < x < \left( 2n + \frac{3}{2} \right) \pi and is decreasing if \left( 2n - \frac{1}{2} \right) \pi < x < \left( 2n + \frac{1}{2} \right) \pi.

  4. We sketch the curve,

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Use derivatives to sketch the graph of a function

Let

    \[ f(x) = x - \sin x. \]

  1. Find all points such that f'(x) = 0;
  2. Determine the intervals on which f is monotonic by examining the sign of f';
  3. Determine the intervals on which f' is monotonic by examining the sign of f'';
  4. Sketch the graph of f.

  1. We take the derivative,

        \[ f'(x) = 1 - \cos x. \]

    Thus,

        \[ f'(x) = 0 \quad \implies \quad 1 - \cos x = 0 \quad \implies \quad x = 2n \pi. \]

  2. f is increasing for all x since 1 - \cos x \geq 0 for all x.
  3. Taking the second derivative,

        \[ f''(x) = \sin x. \]

    Thus, f' is increasing for 2 n \pi < x < (2n+1) \pi and decreasing for (2n-1) \pi < x < 2n \pi.

  4. We sketch the curve,

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Use derivatives to sketch the graph of a function

Let

    \[ f(x) = \sin^2 x. \]

  1. Find all points such that f'(x) = 0;
  2. Determine the intervals on which f is monotonic by examining the sign of f';
  3. Determine the intervals on which f' is monotonic by examining the sign of f'';
  4. Sketch the graph of f.

  1. We take the derivative,

        \[ f'(x) = 2 \sin x \cos x = \sin (2x). \]

    Thus,

        \[ f'(x) = 0 \quad \implies \quad \sin (2x) = 0 \quad \implies \quad x = \frac{n \pi}{2} \]

    for n an integer. (We found the zeros of sine in this exercise, Apostol Section 2.8 Exercise #1.)

  2. f is increasing if n \pi < x < \left( n + \frac{1}{2} \right) \pi and decreasing if \left( n - \frac{1}{2} \right) \pi < x < n \pi.
  3. Taking the second derivative,

        \[ f''(x) = -2 \cos (2x). \]

    Thus, f' is increasing if \left( n - \frac{1}{4} \right) \pi < x < \left( n + \frac{1}{4} \right) \pi and decreasing if \left( n + \frac{1}{4} \right) \pi < x < \left( n + \frac{3}{4} \right) \pi.

  4. We sketch the curve,

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Use derivatives to sketch the graph of a function

Let

    \[ f(x) = \frac{x^2 - 4}{x^2 - 9}. \]

  1. Find all points such that f'(x) = 0;
  2. Determine the intervals on which f is monotonic by examining the sign of f';
  3. Determine the intervals on which f' is monotonic by examining the sign of f'';
  4. Sketch the graph of f.

  1. We take the derivative,

        \[ f'(x) = \frac{-10x}{(x^2 - 9)^2}. \]

    Thus,

        \[ f'(x) = 0 \quad \implies \quad \frac{-10x}{(x^2 -9)^2} = 0 \quad \implies \quad x = 0. \]

  2. f is increasing if x < -3 or -3 < x < 0 and is decreasing if 0 < x < 3 or x > 3.
  3. Taking the second derivative,

        \[ f''(x) = \frac{30(x^2+3)}{(x^2-9)^3}. \]

    Thus, f' is increasing for x < -3 or x > 3 and is decreasing if -3 < x < 3.

  4. We sketch the curve,

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Use derivatives to sketch the graph of a function

Let

    \[ f(x) = \frac{x}{1+x^2}. \]

  1. Find all points such that f'(x) = 0;
  2. Determine the intervals on which f is monotonic by examining the sign of f';
  3. Determine the intervals on which f' is monotonic by examining the sign of f'';
  4. Sketch the graph of f.

  1. We take the derivative,

        \[ f'(x) = \frac{1-x^2}{(1+x^2)^2}. \]

    Thus,

        \[ f'(x) = 0 \quad \implies \quad \frac{1-x^2}{(1+x^2)^2} = 0 \quad \implies \quad x = \pm 1. \]

  2. f is increasing if |x| < 1 and decreasing if |x| > 1.
  3. Taking the second derivative,

        \[ f''(x) = \frac{2x(x^2 - 3)}{(1+x^2)^3}. \]

    Thus, f' is increasing if -\sqrt{3} < x < 0 or x > \sqrt{3} and f' is decreasing if x < - \sqrt{3} or 0 < x < \sqrt{3}.

  4. We sketch the curve,

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Use derivatives to sketch the graph of a function

Let

    \[ f(x) = \frac{1}{(x-1)(x-3)}. \]

  1. Find all points such that f'(x) = 0;
  2. Determine the intervals on which f is monotonic by examining the sign of f';
  3. Determine the intervals on which f' is monotonic by examining the sign of f'';
  4. Sketch the graph of f.

  1. We take the derivative,

        \[ f'(x) = \frac{-2x + 4}{(x-1)^2 (x-3)^2}. \]

    Thus,

        \[ f'(x) = 0 \quad \implies \quad \frac{-2x+4}{(x-1)^2(x-3)^2} = 0 \quad \implies \quad x = 2. \]

  2. f is increasing if x < 1 or 1 < x < 2 and f is decreasing if 2 < x < 3 or x > 3. (We have to take some care here to leave out the points x = 1 and x = 3 since the function is not defined at these points.)
  3. Taking the second derivative,

        \[ f''(x) = \frac{1}{(x-3)^3} - \frac{1}{(x-1)^3}. \]

    Thus, f' is increasing for x < 1 and x > 3 and f' is decreasing for 1 < x < 3.

  4. We sketch the curve,

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Use derivatives to sketch the graph of a function

Let

    \[ f(x) = x + \frac{1}{x^2}. \]

  1. Find all points such that f'(x) = 0;
  2. Determine the intervals on which f is monotonic by examining the sign of f';
  3. Determine the intervals on which f' is monotonic by examining the sign of f'';
  4. Sketch the graph of f.

  1. We take the derivative,

        \[ f'(x) = 1 - \frac{2}{x^3}. \]

    Thus,

        \[ f'(x) = 0 \quad \implies \quad 1 - \frac{2}{x^3} = 0 \quad \implies \quad x = 2^{\frac{1}{3}}. \]

  2. f is increasing if x < 0 or x > 2^{\frac{1}{3}}, and f is decreasing if 0 < x < 2^{\frac{1}{3}}.
  3. Taking the second derivative,

        \[ f''(x) = \frac{6}{x^4}. \]

    Thus, f' is increasing for all x \neq 0 (and is undefined if x = 0).

  4. We sketch the curve,

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Use derivatives to sketch the graph of a function

Let

    \[ f(x) = \frac{1}{x^2}. \]

  1. Find all points such that f'(x) = 0;
  2. Determine the intervals on which f is monotonic by examining the sign of f';
  3. Determine the intervals on which f' is monotonic by examining the sign of f'';
  4. Sketch the graph of f.

  1. We take the derivative,

        \[ f'(x) = - \frac{2}{x^3}. \]

    Thus,

        \[ f'(x) = 0 \quad \implies \quad -\frac{2}{x^3} = 0 \quad \implies \quad f'(x) \text{ is never zero}. \]

  2. f is increasing for x < 0, and decreasing for x > 0.
  3. Taking the second derivative,

        \[ f'(x) = -\frac{2}{x^3} \quad \implies \quad f''(x) = \frac{6}{x^4}. \]

    Thus, f' is increasing for all x \neq 0 (and is undefined at x = 0).

  4. We sketch the curve,

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Use derivatives to sketch the graph of a function

Let

    \[ f(x) = 2 + (x-1)^4. \]

  1. Find all points such that f'(x) = 0;
  2. Determine the intervals on which f is monotonic by examining the sign of f';
  3. Determine the intervals on which f' is monotonic by examining the sign of f'';
  4. Sketch the graph of f.

  1. We take the derivative,

        \[ f'(x) = 4(x-1)^3. \]

    Thus,

        \[ f'(x) = 0 \quad \implies \quad 4(x-1)^3 = 0 \quad \implies \quad x = 1. \]

  2. f is increasing if x > 1 and f is decreasing if x < 1.
  3. Taking the second derivative,

        \[ f'(x) = 4(x-1)^3 \quad \implies \quad f''(x) = 12(x-1)^2. \]

    Thus, f' is increasing for all x since f'' is positive for all x.

  4. We sketch the curve,

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