Question

Find Δf and df for the function f for the indicated values of x, Δx and compare:

f(x) = x² + 2x + 3, x = – 0.5, Δx = dx = 0.1

Answer 1

y = f(x) = x² + 2x + 3

dy = (2x + 2)dx

dy (when x = – 0.5 and dx = 0.1)

= [2(– 0.5) + 2](0.1)

= (– 1 + 2)(0.1)

= (1)(0.1)

= 0.1

(i.e.,) df = 0.1

Now ∆f = f(x + ∆x) – f(x)

Here x = – 0.5 and ∆x = 0.1

x² + 2x + 3

f(x + ∆x) = f(– 0.5 + 0.1)

= f(– 0.4)

= (– 0.4)² + 2(– 0.4) + 3

= 0.16 – 0.8 + 3

= 3.16 – 0.8

= 2.36

f(x) = f(– 0.5)

= (– 0.5)² + 2(– 0.5) + 3

= 0.25 – 1 + 3

= 3.25 – 1

= 2.25

So ∆f = f(x + ∆x) – f(x)

= 2.36 – 2.25

= 0.11