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Find functions satisfying given conditions

by RoRi

Find functions satisfying the given conditions in each of the following cases.

  1. \displaystyle{ f'(x^2) = \frac{1}{x} } for x> 0 and f(1) = 1.
  2. \displaystyle{ f'(\sin^2 x) = \cos^2 x} for all x, and f(1) = 1.
  3. \displaystyle{ f'(\sin x) = \cos^2 x} for all x and f(1) = 1.
  4. \displaystyle{ f'(\log x) = \begin{cases} 1 & \text{for } 0 < x \leq 1, \\ x & \text{for } x > 1, \end{cases}} and f(0) = 0.
  1. We make the substitution t = x^2. Since x > 0 this gives us x = \sqrt{t}. Therefore,

        \begin{align*} f'(x^2) = \frac{1}{x} && \implies && f'(t) &= \frac{1}{\sqrt{t}} \\[9pt] && \implies && \int f'(t) \, dt &= \int \frac{1}{\sqrt{t}} \, dt \\[9pt] && \implies && f(t) &= 2 \sqrt{t} + C. \end{align*}

    Since we are given f(1) = 1 we can solve for C,

        \[ 1 = f(1) = 2 \sqrt{1} + C \quad \implies \quad C = -1. \]

    Therefore,

        \[ f(t) = 2 \sqrt{t} - 1. \]

  2. We make the substitution t = \sin^2 x. Since \sin^2 x + \cos^2 x = 1 we then have \cos^2 x = 1 - t. Therefore,

        \begin{align*} f'(\sin^2 x) = \cos^2 x && \implies && f'(t) &= 1-t \\[9pt] && \implies && \int f'(t) \, dt &= \int (1-t) \, dt \\[9pt] && \implies && f(t) &= t - \frac{t^2}{2} + C. \end{align*}

    Since we are given that f(1) = 1 we can solve for C,

        \[ 1 = f(1) = 1 - \frac{1}{2} + C \quad \implies \quad C = \frac{1}{2}. \]

    Therefore,

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

    This formula is valid for 0 \leq t \leq 1 since t = \sin^2 x and 0 \leq \sin^2 x \leq 1 for all x.

  3. We make the substitution t = \sin x. Since \sin^2 x + \cos^2 x =1 we then have \cos^2 x = 1 - t^2. So,

        \begin{align*} f'(\sin x) = \cos^2 x && \implies && f'(t) &= 1-t^2 \\[9pt] && \implies && \int f'(t) \, dt &= \int (1-t^2) \, dt \\[9pt] && \implies && f(t) &= t - \frac{t^3}{3} + C. \end{align*}

    Since we are given that f(1) = 1 we can solve for C,

        \[ 1 = f(1) = 1 - \frac{1}{3} + C \quad \implies \quad C = \frac{1}{3}. \]

    Therefore,

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

    This is valid for |t| \leq 1 since t = \sin x and -1 \leq \sin x \leq 1 for all x.

  4. We make the substitution t = \log x. Then, x = e^t and so we have

        \[ f'(\log x) = \begin{cases} 1 & \text{for } 0 < x \leq 1, \\ x & \text{for } x > 1, \end{cases} \quad \implies \quad f'(t) = \begin{cases} 1 & \text{for } 0 < x \leq 1, \\ e^t & \text{for } x > 1. \end{cases}\]

    So, we consider the two cases separately. If 0 < x \leq 1 then we have t < 0 and

        \[ f'(t) = 1 \quad \implies \quad \int f'(t) \, dt = \int 1 \, dt \quad \implies \quad f(t) = t + C. \]

    If x > 1 then we have t \geq 0 and

        \[ f'(t) = e^t \quad \implies \quad \int f'(t) \,dt = \int e^t \, dt \quad \implies \quad f(t) = e^t + C. \]

    Therefore, we have the function

        \[ f(t) = \begin{cases} t + C & \text{if } t < 0, \\ e^t + C & \text{if } t \geq 0. \end{cases} \]

    Now, to solve for C we use the condition that f(0) = 0. (Here we’re going to assume we want to make the function continuous at t = 0, i.e., that the two pieces of this piecewise definition take the same value at 0 so that the limits from the left and right would be equal.) Therefore, we have

        \[ 0 = f(0) = 0 + C \quad \implies \quad C = 0 \]

    and

        \[ 0 = f(0) = e^0 + C \quad \implies \quad C = -1.\]

    Thus, the function is given by

        \[ f(t) = \begin{cases} t & \text{if } t < 0, \\ e^t - 1 & \text{if } t \geq 0. \end{cases} \]

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