Ex 6.4,2 - Chapter 6 Class 12 Application of Derivatives

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Ex 6.4,2 Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2. Let x = 2 and x = 0.01 Given f(x) = 4x2 + 5x + 2 f (x) = 8x + 5 Now, = f (x) x = (8x + 5) 0.01 Also, y = f (x + x) f(x) f(x + x) = f (x) + f (2.01) = 4x2 + 5x + 2 + (8x + 5)(0.01) Putting value of x, & 2.01 =4 2 2 + 5 2 +2+ 0.01 8 2 +5 = 16+10+2 + 21 0.01 = 28+0.21=28.21 Hence, the approximate value of f (2.01) is 28.21