# What is the extreme value of f(x)=x^4+8x^2-48x+19 for 1<x<2? Is this a mininum or maximum?

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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.**

Let f(x) x^4+8x^3-48x+19 = 0. To find the extreme value.

The maximum or minimum is for the value of x for which f'(x) = 0.

f'(x) = (x^4+8x^2-48x+19)' = 0.

=> 4x^3+16x-48) = 0.

We divide the equation by 4:

x^3 +4x-12 = 0.

Let g(x) = x^3+4x-12.

g(1) = -12 and g(2) = 2^3+4*2-12 = 4.

Therefore there is a root x = c between 1 and 2 for which g(c) = 0.

g(1) = -12

g(2) = 4.

So c1 = 1+ (2-1)(0-(-12)))/(4-(-12)) = 1.75.

We substitute in c1 in x^3 = (12-4x), Or c2= (12-4*c1)^(1/3) = (12-4*1.75)^(1/3) = 1.71.

C3 = (12-4*C2)^(1/3) = 1.73

C4 = (12-c3)^(1/3) = 1.72

C5 = 12-1.72)^(1/3) = 1.72.

Since C4 = C5, we find that 1.72 is the nearest solution for x^3+4x-12 = 0 and so it must be the estimate for the extreme for f(x) x^4+8x^2-48x+19.

Also f"(x) = (4x^3+16x-48)' = 12x^2+16.

So f''(C5) = 12*(1.72)^2+16> 0.

So f(1.72) is the minimum locally.

Therefore f(1.72) = (1.72)^4+8(1.72)^2-48(1.72) +19 = -31.14 is the local minimum value for x = 1.72.

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