Q. 115.0( 1 Vote )

# Find the approximate value of f(5.001), where f(x) = x^{3} – 7x^{2} + 15.

Answer :

Given f(x) = x^{3} – 7x^{2} + 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) = 5^{3} – 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

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