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

**Solution:**

We can use differentials to calculate small changes in the dependent variable of a function corresponding to small changes in the independent variable

Let x = 5 and Δx = 0.001

Then,

f (5.001) = f (x + Δx)

Replacing x with x + Δx in the given function.

= ( x + Δx)^{3} - 7 (x + Δx)^{2} + 15

Δy = f (x + Δx) - f (x)

f (x + Δx) = f (x) + Δy

≈ f (x) + f' (x).Δx

(∵ dx = Δx)

f (5.001) ≈ (x^{3} - 7x^{2} + 15) + (3x^{2} - 14x)Δx

= [(5)^{3} - 7 (5)^{2} + 15] + [3(5)^{2} - 14 (5)] (0.001)

[∵ x = 5, Δx = 0.001]

= (125 - 175 + 15) + (75 - 70)(0.001)

= (- 35) + (5)(0.001)

= - 35 + 0.005

= - 34.995

Hence, the approximate value of f (5.001) = - 34.995

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.4 Question 3

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

**Summary:**

The approximate value of f (5.001), where f (x) = x^{3} - 7x^{2} + 15 is - 34.995. We can use differentials to calculate small changes in the dependent variable of a function corresponding to small changes in the independent variable

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