Graph the function to get an idea of what you have.
Google “desmos graphing online” and enter the function.
It is a cubic that goes up to a max at (-1.5, 13.1) and a min at about (1.5, -1.1). To make the cubic only have one x-intercept, we need to add some number b to lift the part that is below the x-axis, above the x-axis.
We want to find the relative minimum.
Take the derivative of f(x) = x^3-7x+6 => f’(x)= 3x^2 – 7.
Solve for x for f(x) = 0
3x^2 – 7 = 0
x^2 = 7/3
x = +/- sqrt(7/3)
We know from the graph that the minimum occurs at + sqrt(7/3), but you can put both numbers back into f(x) to check this out.
f(sqrt(7/3)= (7/3)^(3/2) – 7*sqrt(7/3) + 6 ~= -1.1 this number matches the graph, so I am pretty sure I did not make a calculation mistake.
So the answer is the value of b that makes f(x) + b only have one x-intercept is any value of b > (7/3)^(3/2) – 7*sqrt(7/3) + 6
(Don’t write >= because then the graph will touch the x-axis and there will be two x-intercepts.)