(b)
(c)(d)(a) At (−1,8), so the tangent line is y − 8 = 5(x − (−1)).
Therefore f (x) ≈ 8 + 5(x + 1).
f (3) ≈ 8 + 5(0 + 1) = 13.
At (−1,8), For Δx = 0.5, Δy = 0.5(5) = 2.5, so move to
(−1 + 0.5, 8 + 2.5) = (−0.5,10.5).
At (−0.5,10.5), For Δx = 0.5, Δy = 0.5(8)= 4, so move to (−0.5 + 0.5, 10.5 + 4).
Thus f (0) ≈ 14.5.