Let $f(x) = \frac{3x+2}{x^2-2x+5}$. If $M = 125 f''(0)$ and $N = 625 f'''(0)$, find the value of $M

Question

Let $f(x) = \frac{3x+2}{x^2-2x+5}$. If $M = 125 f''(0)$ and $N = 625 f'''(0)$, find the value of $M - N$.

Professor PC Thomas Classes · Answer

By rearranging the given equation, we get $f(x)(x^2-2x+5) = 3x+2$. Differentiating this identity $n$ times using Leibniz's theorem (for $n \ge 2$): $f^{(n)}(x)(x^2-2x+5) + n f^{(n-1)}(x)(2x-2) + \frac{n(n-1)}{2} f^{(n-2)}(x)(2) = 0$. Substituting $x=0$, we obtain $5f^{(n)}(0) - 2n f^{(n-1)}(0) + n(n-1)f^{(n-2)}(0) = 0$, which simplifies to $f^{(n)}(0) = \frac{2}{5}n f^{(n-1)}(0) - \frac{n(n-1)}{5} f^{(n-2)}(0)$. We first find the initial values: $f(0) = \frac{2}{5}$, and $f'(x) = \frac{3(x^2-2x+5) - (3x+2)(2x-2)}{(x^2-2x+5)^2} \implies f'(0) = \frac{3(5) - (2)(-2)}{25} = \frac{19}{25}$. Now, using the recurrence relation for $n=2$: $f''(0) = \frac{2}{5}(2)f'(0) - \frac{2(1)}{5}f(0) = \frac{4}{5}(\frac{19}{25}) - \frac{2}{5}(\frac{2}{5}) = \frac{76}{125} - \frac{20}{125} = \frac{56}{125}$. Thus, $M = 125 f''(0) = 56$. Next, for $n=3$: $f'''(0) = \frac{2}{5}(3)f''(0) - \frac{3(2)}{5}f'(0) = \frac{6}{5}(\frac{56}{125}) - \frac{6}{5}(\frac{19}{25}) = \frac{336}{625} - \frac{114 	imes 5}{625} = \frac{336 - 570}{625} = -\frac{234}{625}$. Thus, $N = 625 f'''(0) = -234$. Finally, $M - N = 56 - (-234) = 290$.