f(x)=xʸlog(x).
With y constant f'(x)=xʸ/x+yxʸ⁻¹log(x)=xʸ⁻¹+yxʸ⁻¹log(x).
Rolle’s Theorem requires continuity of f(x) in the interval [0,1] and continuity of f'(x) in the open interval (0,1).
f(x)=xʸlog(1-(1-x)). f(x)=xʸ(-(1-x)-(1-x)²/2-(1-x)³/3-...)=-xʸ((1-x)+(1-x)²/2+...). x→0, f(x)→0, provided y>0.
f'(x) is continuous in (0,1).
So Rolle’s Theorem requires y>0.
f(0)=f(1)=0, f'(x)=0=xʸ⁻¹+yxʸ⁻¹log(x)=xʸ⁻¹(1+ylog(x)), so ylog(x)=-1, log(x)=-1/y, x=e^(-1/y) where the log base is assumed to be e, natural log base.
The graph shows f(x) and f'(x) for various values of y:
Red, y=0.25
Green, y=0.5
Black, y=0.75
Blue, y=1
Purple, y=2
Orange, y=3
f(x)≤0 for x in [0,1] (on or below the x-axis), while f'(x) has both positive and negative values.
Where f'(x)=0, f(x) is minimum.