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Prove that the function f(x)=x4−4x3+6x2−4x+1 has no real roots.
Jul 3, 2025
Prove that the function f(x)=x4−4x3+6x2−4x+1 has no real roots.
Solution by Steps
step 1
To determine if the function f(x)=x4−4x3+6x2−4x+1 has real roots, we can analyze its derivative to find critical points
step 2
The derivative is given by f′(x)=4x3−12x2+12x−4. We can factor this to find critical points
step 3
Factoring gives us f′(x)=4(x3−3x2+3x−1). We can use the Rational Root Theorem to test for possible rational roots
step 4
Testing possible rational roots shows that x=1 is a root of f′(x), leading to f′(1)=0. We can use this to find the nature of the critical point
step 5
Evaluating f(1)=14−4(1)3+6(1)2−4(1)+1=0, indicates that x=1 is a point of interest
step 6
To confirm that f(x) has no real roots, we can analyze the second derivative f′′(x)=12x2−24x+12. Evaluating f′′(1)=0 suggests a possible inflection point
step 7
The function f(x) can be rewritten as (x−1)4, which is always non-negative and equals zero only at x=1. Thus, it has no real roots
Answer
The function f(x)=x4−4x3+6x2−4x+1 has no real roots, as it is non-negative and equals zero only at x=1.
Key Concept
The nature of polynomial functions and their roots can be analyzed using derivatives and factoring.
Explanation
The analysis shows that the function does not cross the x-axis, confirming it has no real roots. The only point where it equals zero is a repeated root.
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