The value of $$6m_{2} + 4n_{2} + 8m_{2}n_{2}$$ is ___ .

Write $$f_2(x) = 98(x-1)^{50}-600(x-1)^{49}+2450,\; x\gt 0$$ and set

$$y = x-1 \;\Longrightarrow\; y \in (-1,\infty).$$

In terms of $$y$$, $$f_2(x)=g(y) = 98y^{50}-600y^{49}+2450.$$ Finding local extrema reduces to analysing $$g(y)$$ on $$(-1,\infty).$$

Step 1: First derivative
$$g'(y)=\frac{d}{dy}\bigl(98y^{50}-600y^{49}+2450\bigr) =98\cdot 50\,y^{49}-600\cdot 49\,y^{48}$$ $$\;\;\;=4900y^{49}-29400y^{48} =4900y^{48}(y-6).$$

Step 2: Stationary points
Set $$g'(y)=0.$$
$$4900y^{48}(y-6)=0 \;\Longrightarrow\; y=0\;(\text{multiplicity }48)\quad\text{or}\quad y=6.$$
Thus potential extrema in $$x$$-notation are

$$x=1\;(y=0)\quad\text{and}\quad x=7\;(y=6).$$

Step 3: Nature of each stationary point

• Around $$y=0$$:
Pick $$y=-0.1$$ (still >-1): $$g'(y)=4900(-0.1)^{48}\bigl(-0.1-6\bigr)\lt 0.$$
Pick $$y=+0.1$$: $$g'(y)=4900(0.1)^{48}\bigl(0.1-6\bigr)\lt 0.$$
Since $$g'$$ is negative on both sides and merely touches $$0$$ at $$y=0$$, there is no change of sign; hence $$x=1$$ is neither a local minimum nor a local maximum.

• Around $$y=6$$:
For $$y=5.9$$, $$y^{48}\gt 0$$ but $$(y-6)\lt 0$$ ⇒ $$g'(5.9)\lt 0.$$ For $$y=6.1$$, $$y^{48}\gt 0$$ and $$(y-6)\gt 0$$ ⇒ $$g'(6.1)\gt 0.$$ Derivative changes from negative to positive, so $$x=7$$ is a point of local minimum.

Step 4: Counting extrema for $$f_2$$ on $$(0,\infty)$$
Number of local minima: $$m_2 = 1\;(x=7).$$
Number of local maxima: $$n_2 = 0.$$

Step 5: Required expression
$$6m_2 + 4n_2 + 8m_2n_2 = 6(1) + 4(0) + 8(1)(0) = 6.$$

Hence the value is $$6,$$ which lies in the indicated range $$5.90\text{-}6.10.$$