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Mathematics-Online problems:

Solution to the problem of the (previous) week

Problem:

The function

$\displaystyle f(x) = 1 + \frac{a-bx}{x^3} $

has a double zero at $ x = 1$ .


\includegraphics[width=.5\linewidth]{TdM_1994_3}


(i)     Determine the parameters $ a$ and $ b$ .
(ii) Determine all zeros of $ f$ .
(iii) Find the area $ A_c$ of the shaded region for $ c = 2$ as well as $ \displaystyle\lim_{c\to\infty} A_c$ .
(iv) Sketch the graph for $ x < 0$ .


Answer:

(i)     Parameters $ a=$ $ ,$ $ b=$
(ii) Another zero $ z=$
(iii) Area $ A_c=$ $ ,$     limiting value $ \displaystyle\lim_{c\to\infty} A_c=$
(iv) The graph of the function for $ x < 0$ is:    not specified

\includegraphics[bb=140 345 481 620,clip,width=.35\linewidth]{TdM_1994_3_2} \includegraphics[bb=140 345 480 630,clip,width=.35\linewidth]{TdM_1994_3_4}
\includegraphics[bb=140 345 480 630,clip,width=.35\linewidth]{TdM_1994_3_3} \includegraphics[bb=140 345 480 620,clip,width=.35\linewidth]{TdM_1994_3_1}

Solution:

(i)
Determination of the parameters $ a$ and $ b$:
    $\displaystyle f(1) = 0 \quad\Rightarrow\quad 1 + a - b = 0 \quad\Rightarrow\quad b = a + 1$  
    $\displaystyle f'(x) = -\frac{b}{x^3} - 3\frac{a-bx}{x^4}$  
    $\displaystyle f'(1) = 0 \Rightarrow -(a+1)-3\cdot (-1) = 0 \quad\Rightarrow\quad a = 2$  
    $\displaystyle f(x) = 1 + \frac{2-3x}{x^3}.$  

(ii)
Besides the double zero at $ x = 1$, $ f$ has one more zero $ z$:

$\displaystyle 0 = x^3 - 3x + 2 = (x-1)^2(x-z)\quad \Rightarrow\quad z = -2. $

(iii)
$ f(x) = 1 \Rightarrow x = 2/3$. This yields the area $ A_c$:
    $\displaystyle A_c = \int_{2/3}^c \frac{3x-2}{x^3}\,dx = \left[-\frac{3}{x} + \f... ...2}\right]_{2/3}^c = \frac{9}{4} + \left(\frac{1}{c^2} - \frac{3}{c}\right) = 1;$  
    $\displaystyle \lim_{c\to\infty} A_c = \frac{9}{4} = 2.25.$  

(iv)
The graph of the function for $ x < 0$:


\includegraphics[width=0.5\textwidth]{TdM_1994_3_1}
[problem of the week]