Let,
y=x4+147x3=f(x)
We know that, for increasing function dxdy>0
So, differentiating the given function f(x)=x4+147x3 we get,
dxdy=(x4+147)2−x2(x4−441)
⇒(x4+147)2−x2(x4−441)>0
⇒(x4+147)2x2(x4−441)<0
⇒x4<441
Now for maxima/minima dxdy=0
⇒x4=441
⇒x=α,4<α<5
⇒Maximum value of f(x) is at x=4 or x=5
f(4)=40364,f(5)=772125
∵f(5)>f(4)
⇒α=5