To find the extreme value of a function f(x) we need to find the first derivative of the function and equate it to zero. Then this is used to solve for x.

Here, f(x)=x^4 + 8x^2 - 48x + 19

f'(x) = 4x^3 + 16x - 48

4x^3 + 16x - 48 = 0

=> x^3 + 4x - 12 = 0

This equation has one real root which is 1.7224, and it lies between 1 and 2

f( 1.7224) = -31.14

At x = 1.7224 f''(x) = 12x^2 - 16 = 19.59 which is positive.

**Therefore the minimum value of the function f(x)=x^4 + 8x^2 - 48x + 19 for 1<x<2 is -31.14.**

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