<s>[INST] At which points does the graph of the $f(x)$ have horizontal tangent line? $(2x^2-2x-1)/(x^4+1)$
I tried to solve with the derivative method. But came across a fifth degree Polynomial involving $5$ terms. After that my math failed to factorize it. This is the derivative $2x^5 - 3x^4-2x^3-2x+1$. Kindly help.
 [/INST] With derivative method, we get
$$f'(x) = \frac{(4x-2)(x^4+1)-4x^3(2x^2-2x-1)}{(x^4+1)^2} = 0.$$
We have to find $x$ so that the equality can be fulfilled.
We obtain
$$(4x-2)(x^4+1) - 4x^3(2x^2-2x-1) = 0$$
$$ \iff (4x^5-2x^4 + 4x-2)- 8x^5+8x^4+4x^3 = 0 $$
$$\iff -4x^5 + 6x^4+4x^3+4x-2 = 0 $$
$$\iff 2x^5 - 3x^4 -2x^3-2x+1 = 0.$$
Define $g(x) = 2x^5 - 3x^4 -2x^3-2x+1$.
If you substitute $x = -1$, then $g(-1) = 0$. So $(x+1)$ is the factor of $g(x)$.
We get
$$g(x) = (x+1)(2x^4-5x^3+3x^2-3x+1).$$
From this equation, we obtain at $x=-1$, $f(x)$ has a horizontal tangent line.
You can find 2 more solution (if I'm not mistaken).
Hint: Find $x$ as a solution of $2x^4-5x^3+3x^2-3x+1 = 0.$
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