Evaluation of Local Extrema Expression for Second Polynomial Function

To find the value of $6m_2 + 4n_2 + 8m_2 n_2$, we need to determine the number of points of local minima ($m_2$) and local maxima ($n_2$) for the function $f_2(x)$ in the interval $(0, \infty)$.

The given function is: $f_2(x) = 98(x - 1)^{50} - 600(x - 1)^{49} + 2450, \quad \text{for } x > 0$

Differentiating $f_2(x)$ with respect to $x$: $f_2'(x) = 98 \cdot 50 (x - 1)^{49} - 600 \cdot 49 (x - 1)^{48}$

Simplifying the coefficients: $98 \cdot 50 = 4900$ $600 \cdot 49 = 29400$

So, the derivative becomes: $f_2'(x) = 4900 (x - 1)^{49} - 29400 (x - 1)^{48}$

Factoring out $4900 (x - 1)^{48}$: $f_2'(x) = 4900 (x - 1)^{48} \left[ (x - 1) - 6 \right]$ $f_2'(x) = 4900 (x - 1)^{48} (x - 7)$

Now, let's analyze the critical points of $f_2(x)$ in $(0, \infty)$, which are $x = 1$ and $x = 7$:

1. At $x = 1$:

   • For $x \in (0, 1)$, $(x - 1)^{48} > 0$ and $(x - 7) < 0$, so $f_2'(x) < 0$.
   • For $x \in (1, 7)$, $(x - 1)^{48} > 0$ and $(x - 7) < 0$, so $f_2'(x) < 0$. Since $f_2'(x)$ does not change sign around $x = 1$, the point $x = 1$ is neither a local minimum nor a local maximum (it is a point of inflection).

2. At $x = 7$:

   • For $x \in (1, 7)$, $f_2'(x) < 0$.
   • For $x \in (7, \infty)$, $(x - 1)^{48} > 0$ and $(x - 7) > 0$, so $f_2'(x) > 0$. Since $f_2'(x)$ changes sign from negative to positive as $x$ passes through $7$, $x = 7$ is a point of local minimum.

Thus, for $f_2(x)$ in the interval $(0, \infty)$:

• Number of points of local minima, $m_2 = 1$
• Number of points of local maxima, $n_2 = 0$

Substitute the values of $m_2$ and $n_2$ into the required expression: $6m_2 + 4n_2 + 8m_2 n_2 = 6(1) + 4(0) + 8(1)(0) = 6$