Question 317255
{{{f(1)=2(1)^5-7(1)+1=2-7+1=-4}}}
{{{f(2)=2(2)^5-7(2)+1=51}}}
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Since {{{-4 < 0 < 51}}}, then by the intermediate value theorem, there is an {{{x}}}, ({{{1 < x < 2}}}), such that {{{f(x)=0}}}.