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Find a function whose graph has given properties

Let f be a function such that the points P_0 = (0,1) and P_1 = (1,0) are on the graph of f. Assume that for every point P = (x,y) on the graph of f, the graph is above the line segment joining P_0 to P. Further, assume that the area A(x) of the region bounded by the graph of f and the line segment is x^3. Find a formula for the function f.


One comment

  1. Evangelos says:

    From our given information, we wish to find a function f(t) such that the area enclosed by the function and the line segment from (0,1) to (x, f(x)) is always equal to x^3. Since the line segment goes through (0, 1) and (x, f(x)), we can write the equation of the line in point-slope form:

        \begin{align*} f(x) - 1 &= m(x) \end{align*}

    Solving for slope m gives us:

        \begin{align*} m = \frac{f(x) - 1}{x} \end{align*}

    And the equation of the line y = mt + b becomes:

        \begin{align*} y = \frac{f(x) - 1}{x} \ t + 1 \end{align*}

    And since we know that f(t) is always greater than y = mt + b on (0, x), we can write it out the area of the enclosed region through the following integrals:

        \begin{align*} \int_{0}^{x} f(t) dt - [\frac{f(x) - 1}{x} \ t + 1] \ dt & = \int_{0}^{x} f(t) dt - \int_{0}^{x} \frac{f(x) - 1}{x} \ t \ dt - \int_{0}^{x} dt \\ \end{align*}

    Evaluating the polynomial-termed integrals gives us the following

        \begin{align*} \int_{0}^{x} f(t) dt - \frac{f(x) - 1}{2x} \ [t^2]_{0}^{x} - [t]_{0}^{x} &= \int_{0}^{x} f(t) dt -\frac{x}{2}f(x) - \frac{x}{2} \end{align*}

    And from our given information, we know that the area of the enclosed region is always x^3. So:

        \begin{align*} \int_{0}^{x} f(t) dt -\frac{x}{2}(f(x)) - \frac{x}{2} &= x^3 \end{align*}

    Differentiating both sides with respect to x gives us:

        \begin{align*} \frac{1}{2}(f(x)) - \frac{x}{2}(f'(x)) - \frac{1}{2} &= 3x^{2} \end{align*}

        \begin{align*} f(x) - x(f'(x)) - 1 &= 6x^{2} \end{align*}

    Now, since the right-hand side is a second degree polynomial, it may be the case that f(x) is one as well. If f(x) is a second-degree polynomial

        \begin{align*} f(x) = Ax^2 + Bx + C \end{align*}

    Then, its derivative is

        \begin{align*} f'(x) = 2Ax + B \end{align*}


        \begin{align*} f(x) - x(f'(x)) - 1 &= -Ax^2 + C - 1 \\ &= 6x^{2} \end{align*}

    Solving for A and C, we get:

        \begin{align*} A = -6 \\ C = 1 \end{align*}

    And using our given that f(1) = 0, we can solve for B

        \begin{align*} -6 + B + 1 = 0\\ B = 5 \end{align*}

    And our polynomial f(x) is

        \begin{align*} f(x) = -6x^2 + 5x + 1 \end{align*}

    To confirm that this is indeed the function that satisfies all the initial conditions, it should suffice to evaluate the original integrals using f(t) = -6t^2 + 5t + 1. Or, to check the back of the book and see that it matches :)

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