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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 $p$ .