Calculus Tangent slope ?

No derivatives nor limits eh? Normally I'd suggest at least a difference quotient but that still requires a limit. So you'd have to do it numerically.

F(x) = sqrt(x)

F(4) = 2

So We want the slope around it, so lets make a table:

F(3.5) = 1.87083

F(3.7) = 1.92354

F(3.9) = 1.97484

F(4.1) = 2.02485

F(4.3) = 2.07364

F(4.5) = 2.12132

So lets calculate some slopes now:

m₁ = (F(3.5) - F(4))/(3.5 - 4) = (1.87083 - 2)/(3.5 - 4) = 0.258343

m₂ = 0.254872

m₃ = 0.251582

m₄ = 0.248457

m₅ = 0.245480

m₆ = 0.242641

We have a bunch of numeric slopes centered at the point of interest, now we can average them to get a good idea as to what the slope should be.

1/6 (Σ m) = 0.250229

So our slope at that point is: 0.250229