We have:
(X-11)/x^4 = 2 - 4/x
Let's pass x^4 to the right side of equation
(X-11) = x^4.(2 - 4/x)
Use the disttibutive
x - 11 = 2x^4 - 4x^4 / x
Look:
4x^4 / x = 4x^3
Then,
X - 11 = 2x^4 -4x^3
2x^4 - 4x^3 - x + 11 = 0
Let's to use the derivative method:
Let f(x) = 2x^4 - 4x^3 - x +11
Then,
f(x)' = 2.(4).x^3 - 4.(3).x^2 -1
f(x)' = 8x^3 -12x^2 -1
The root to the equation is aproximately X = 1,56
This value would be the minumum or local maximum.
Replacing in the 2 derivative the value of X = 1,56
f(x)'' = 24x^2 - 24x
f(x)'' = 24(1,56)^2 - 24.(1,56)
f(x)" = 20,96
Second tge rule of derivative
If f(x)" > 0
X is point of minumum
Or
If f(x)" < 0
X is point of maximum
Then,
X = minumum in 1,56
Now, let's to replace the value of X = 1,56 in the original equation
f(x) = 2x^4 - 4x^3 - x +11
f(x) = 2.(1,56)^4 - 4.(1,56)^3 -1,56 +11
f(x) ~ 6, 1
We know too:
11 It's where it cuts the axis "Y"
Look the picture here down,
There is not solution to this question!
I hope this has helped!
