Use the IVT to prove that y=xlnx has at least one zero on (1/2 , 3/2)

Answer 1

f(1/2) = about -.35

f(3/2) = about +.41

Since it's a continuous function. the line must cross the x axis at least once between x=1/2 and x=3/2

f(1) = xlnx = (1)ln(1) = 0 x=1 is the root or zero

Answer 2

Well for CONTINUOUS function f(x) = x * lnx on (1/2,3/2),

f(1/2) < 0 and f(3/2)>0

so IVT guarantees there is c in (1/2,3/2) where f(c)=0