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Answer 1 · 2018-05-21T17:10:55

Please see below.

#f(x) = 1/27(ax-1)^3(b-3x)+1#

#f'(x) = 1/27[3a(ax-1)^2(b-3x)-3(ax-1)^3]#

# = 1/27 * 3(ax-1)^2[a(b-3x)-(ax-1)]#

# = 1/9 (ax-1)^2(ab-4ax+1)#

#f'(x) = 0# at #x= 1/a# and at #x=(ab+1)/(4a)#

Both solutions require #a# in a denominator, so #f# has no stationary points if #a = 0#. (Observe that in that case #f(x)# is linear.)

#f(1/a) = 1#, so one stationary point is #(1/a,1)#.

If #(1,1)# is a stationary point and there is another #(p,p)# with #p != 1#, we must have #1/a = 1# so, #a =1#

The other stationary point is, therefore, #p = (b+1)/4#.

We are told #f(p) = p#, so

Set #f((b+1)/4)= (b+1)/4# and solve for #b#, then find #p#.