The function f(x)=x5−x3+3x−5 is continuus on R as a polynomial.
Then the function f is continuous on the closed interval [1,2].
f(1)=(1)5−(1)3+3(1)−5=−2<0
f(2)=(2)5−(2)3+3(2)−5=25>0
f(1)=−2<0<25<f(2) Hence the Intermediate Value Theorem can be applied.
Then there exists a number c in (0,1) such that f(c)=0.
Therefore the equation x5−x3+3x−5=0 has at least one root c in the interval (1,2).

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