Estimate f′(6) where f(x)=x3+4x2+4x−18 to within one decimal place by using a small enough interval.
f′(6)≈
let's try x = 6.01 and x = 6
f(6.01) = 367.6022
f(6) = 366
f′(6)≈ (367.6022 - 366)/(6.01-6) ≈ 160.22
actual answer:
f'(x) = 3x^2 + 8x + 4
f'(6) = 216 + 48 + 4 = 160
trying something like x = 6.0001 will bring us much closer to 160, after
all, we got calculators, right?