f(x)=x5+4x−1 The function f(x) is continuous on R as polynomial.
f(0)=0+0−1=−1<0
f(1)=1+4−1=4>0 By the Intermediate Value Theorem there exists a number c∈(0,1) such that f(c)=0.
f′(x)=5x4+4,Since f′(x)>0 for x≥0, the function is strictly increasing on [0,∞). Hence there is the only number c∈(0,1) such that f(c)=0.
Therefore the equation x5+4x=1 has exactly one solution on [0,1].
Comments