Prove that the function $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$ has no real roots.

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∻Solution by Steps∻ ‖ step 1 ⋮ To prove that the function $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$ has no real roots, we can analyze the function by checking its discriminant or by evaluating its behavior. ‖ ‖ step 2 ⋮ We can rewrite the function as $f(x) = (x-1)^4$, which is a perfect square. ‖ ‖ step 3 ⋮ Since $(x-1)^4 \geq 0$ for all real $x$, the function $f(x)$ is always non-negative. ‖ ‖ step 4 ⋮ The only point where $f(x) = 0$ is at $x = 1$, which is a double root. Thus, there are no real roots where $f(x)$ changes sign. ‖ ∻Answer∻ ⚹ The function $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$ has no real roots except at $x = 1$, which is a double root. ⚹ ∻Key Concept∻ ⚹ A polynomial function can have real roots where it changes sign, and analyzing its structure can reveal the nature of its roots. ⚹ ∻Explanation∻ ⚹ The function is always non-negative and only touches the x-axis at $x = 1$, indicating it does not have distinct real roots. ⚹◊What is the derivative of the function $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$, and how does it help in analyzing the function's behavior?⍭ Generate me a similar question◊

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