Use linear approximation, i.e. the tangent line, to approximate (1/0.252) as follows Let f(x)= 1/x and find the equation of the tangent line in slope-intercept form to f(x) at a "nice" point near0.252

Question

Then use this to approximate (1/0.252)

Answer 1

f(.25) = 4

f^'(x) = -1/x²

f^'(.25) = - 16

Eqn of tangent (in pt-slope form): y - 4 = - 16(x - .25). in slope-int form : y = -16x + 8

y(.252) = 3.968

∴ f(.252) ≈ 3.968