What is the solution set of the equation? x-11/x 4 =2-4/x?

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!

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