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Tamil Nadu Board of Secondary EducationHSC Science Class 12

Find Δf and df for the function f for the indicated values of x, Δx and compare: f(x) = x3 – 2x2, x = 2, Δx = dx = 0.5

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Question

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

f(x) = x3 – 2x2, x = 2, Δx = dx = 0.5

Sum
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Solution

y = f(x) = x3 – 2x2

dy = (3x2 – 4x)dx

dy (when x = 2 and dx = 0.5) = [3(22) – 4(2)](0.5)

= (12 – 8)(0.5)

= 4(0.5)

= 2

(i.e.,) df = 2

Now ∆f = f(x + ∆x) – f(x)

Here x = 2 and ∆x = 0.5

f(x) = x3 – 2x2

So f(x + ∆x) = f(2 + 0.5)

= f(2.5) = (2.5)3 – (2.5)2 

= (2.5)2 [2.5 – 2]

= 6.25(0.5)

= 3.125

f(x) = f(2) = 23 – 2(22)

= 8 – 8

= 0

So ∆f = 3.125 – 0

= 3.125

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Linear Approximation and Differentials
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Chapter 8: Differentials and Partial Derivatives - Exercise 8.2 [Page 67]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 8 Differentials and Partial Derivatives
Exercise 8.2 | Q 3. (i) | Page 67

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