If a function f(a) = max (a, 0) then the smallest integer value of x for which the equation f(x — 3) + 2𝑓(x + 1) = 8 holds true is _______________.

f(a) = max (a, 0). This means that is a>0, then f(a) = a, and if a<0, then f(a) = 0.

We are given - f(x — 3) + 2𝑓(x + 1) = 8

=> [max (x-3, 0)] + 2[max (x+1, 0)] = 8

The above equation will depend on the range of x. Thus, there are three cases possible - 

Case-1: $$x\ge3$$

f(x-3) = [max (x-3, 0)] = x-3   (Because x-3 will always be greater than or equal to 0 for this case)

f(x+1) = [max (x+1, 0)] = x+1   (Because x+1 will always be greater than or equal to 0 for this case)

Thus, our equation becomes =>   (x-3) + 2(x+1) = 8  => 3x-1 = 8  =>  x = 3.

And, since we assumed that x is greater than or equal to 3, x = 3 satisfies this equation.

Case-2: $$-1\le x<3$$

f(x-3) = [max (x-3, 0)] = 0 (Because x-3 will always be less than 0 for this case, and thus the max value will be 0)

f(x+1) = [max (x+1, 0)] = x+1 (Because x+1 will always be greater than or equal to 0 for this case)

Thus, our equation becomes => 0 + 2(x+1) = 8 => x + 1 = 4 => x = 3.

But, since we assumed that x is less than 3, x = 3 does not satisfy this equation.

Case-3: $$x\le-1$$

f(x-3) = [max (x-3, 0)] = 0 (Because x-3 will always be less than 0 for this case, and thus the max value will be 0)

f(x+1) = [max (x+1, 0)] = 0 (Because x+1 will always be less than 0 for this case)

Thus, our equation becomes => 0 + 0 = 8. But zero is never equal to 8 in any case, thus this does not satisfy the equation. 

After all three cases, we have only one value of x, x = 3, which satisfies the equation. Therefore $$x=3$$.