Asked by bob
how many real solutions does the function f(x)=x^4-4x^2+x+1 have
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Answered by
GPT-5 mini
AI
There are 4 real solutions.
Reason: f is a continuous polynomial. Evaluate at integer points:
- f(-3)=81-36-3+1=43 > 0
- f(-2)=16-16-2+1=-1 < 0 → root in (-3,-2)
- f(-1)=1-4-1+1=-3 < 0
- f(0)=1 > 0 → root in (-1,0)
- f(1)=1-4+1+1=-1 < 0 → root in (0,1)
- f(2)=16-16+2+1=3 > 0 → root in (1,2)
By the Intermediate Value Theorem there is one root in each of the intervals (-3,-2), (-1,0), (0,1), (1,2) — four distinct real roots. A quartic has at most 4 real roots, so this is all of them.
Reason: f is a continuous polynomial. Evaluate at integer points:
- f(-3)=81-36-3+1=43 > 0
- f(-2)=16-16-2+1=-1 < 0 → root in (-3,-2)
- f(-1)=1-4-1+1=-3 < 0
- f(0)=1 > 0 → root in (-1,0)
- f(1)=1-4+1+1=-1 < 0 → root in (0,1)
- f(2)=16-16+2+1=3 > 0 → root in (1,2)
By the Intermediate Value Theorem there is one root in each of the intervals (-3,-2), (-1,0), (0,1), (1,2) — four distinct real roots. A quartic has at most 4 real roots, so this is all of them.
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