We need to find the number of local minima and local maxima of the given function that actually appears in the asked expression.

The question defines both $f_1$ and $f_2$, but the final expression is $6m_2+4n_2+8m_2n_2,$ so only $f_2$ matters.

1. Given function

$f_2(x)=98(x-1)^{50}-600(x-1)^{49}+2450,\qquad x>0.$

Let $y=x-1.$ Then $y>-1$, and $f_2(x)=98y^{50}-600y^{49}+2450.$

To find local minima/maxima, compute the derivative.

2. First derivative

$f_2'(x)=98\cdot 50(x-1)^{49}-600\cdot 49(x-1)^{48}.$

So, $f_2'(x)=4900(x-1)^{49}-29400(x-1)^{48}.$

Factorizing, $f_2'(x)=4900(x-1)^{48}\big((x-1)-6\big).$

Hence, $f_2'(x)=4900(x-1)^{48}(x-7).$

Critical points in $(0,\infty)$ are obtained from $f_2'(x)=0 \implies x=1 \text{ or } x=7.$

3. Sign analysis of $f_2'(x)$

Since $(x-1)^{48}\ge 0$ and is positive for $x\ne 1$, its sign does not change across $x=1$ because the power is even.

Thus the sign of $f_2'(x)$ for $x\ne 1$ is determined by $(x-7)$.

Intervals:

1. For $0<x<1$: $f_2'(x)<0$ because $(x-1)^{48}>0$ and $(x-7)<0$.

2. For $1<x<7$: $f_2'(x)<0$ again because $(x-1)^{48}>0$ and $(x-7)<0$.

3. For $x>7$: $f_2'(x)>0.$

So:

• At $x=1$, derivative is zero but sign remains negative on both sides $Rightarrow$ neither maximum nor minimum.
• At $x=7$, derivative changes from negative to positive $Rightarrow$ local minimum.

Therefore, $m_2=1,\qquad n_2=0.$

4. Required value

$6m_2+4n_2+8m_2n_2=6(1)+4(0)+8(1)(0)=6.$

5. Comparison with stored answer

Derived answer = $6$.

Stored correct answer = $6.00$.

They match.