Find the approximate value of f(5.001), where f(x) = x3 – 7x2 + 15.
Given f(x) = x3 – 7x2 + 15
Let x = 5 so that x + Δx = 5.001
⇒ 5 + Δx = 5.001
∴ Δx = 0.001
On differentiating f(x) with respect to x, we get
We know and derivative of a constant is 0.
When x = 5, we have
Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as
Here, and Δx = 0.001
⇒ Δf = (5)(0.001)
∴ Δf = 0.005
Now, we have f(5.001) = f(5) + Δf
⇒ f(5.001) = 53 – 7(5)2 + 15 + 0.005
⇒ f(5.001) = 125 – 175 + 15 + 0.005
⇒ f(5.001) = –35 + 0.005
∴ f(5.001) = –34.995
Thus, f(5.001) = –34.995