Question #295391

Show that x^5 +4x=1 has exactly one solution on [0,1]


1
Expert's answer
2022-02-09T13:50:33-0500
f(x)=x5+4x1f(x)=x^5 +4x-1

The function f(x)f(x) is continuous on R\R as polynomial.


f(0)=0+01=1<0f(0)=0+0-1=-1<0

f(1)=1+41=4>0f(1)=1+4-1=4>0

By the Intermediate Value Theorem there exists a number c(0,1)c\in (0, 1) such that f(c)=0.f(c)=0.

f(x)=5x4+4,f'(x)=5x^4+4,

Since f(x)>0f'(x)>0 for x0,x\ge0, the function is strictly increasing on [0,).[0, \infin). Hence there is the only number c(0,1)c\in (0, 1) such that f(c)=0.f(c)=0.

Therefore the equation x5+4x=1x^5+4x=1 has exactly one solution on [0,1].[0,1].



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