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तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा १२

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

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प्रश्न

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

योग
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उत्तर

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

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Linear Approximation and Differentials
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Differentials and Partial Derivatives - Exercise 8.2 [पृष्ठ ६७]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 8 Differentials and Partial Derivatives
Exercise 8.2 | Q 3. (ii) | पृष्ठ ६७

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