Example 6 - Chapter 13 Class 11 Limits and Derivatives

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Example ,6 (Method 1) Find the derivative of the function f(x) = 2x2 + 3x – 5 at x = –1. Also prove that f’(0) + 3f’( –1) = 0. Given f(x) = 2x2 + 3x – 5 We know that f’(x) = lim﷮h→0﷯ f﷮ 𝑥 + ℎ﷯ − f (x)﷯﷮h﷯ Now f (x) = 2x2 + 3x – 5 So, f (x + h) = 2(x + h)2 + 3(x + h) – 5 Putting values f’ (x) = lim﷮h→0﷯ (2(𝑥 + ℎ)﷮2﷯ + 3 𝑥 + ℎ﷯ − 5) − (2𝑥﷮2﷯ + 3𝑥 − 5)﷮h﷯ f’ (x) = lim﷮h→0﷯ (2(𝑥 + ℎ)﷮2﷯ + 3 𝑥 + ℎ﷯ − 5) − (2𝑥﷮2﷯ + 3𝑥 − 5)﷮h﷯ Putting x = – 1 f’( –1) = lim﷮h→0﷯ (2(−1+ℎ)﷮2﷯ + 3 −1+ℎ﷯ − 5) − (2 ((−1)﷮2﷯) + 3(−1) − 5)﷮h﷯ = lim﷮h→0﷯ (2(−1+ℎ)﷮2﷯ + 3 −1+ℎ﷯ − 5) − (2 1﷯ − 3 − 5)﷮h﷯ = lim﷮h→0﷯ (2(−1+ℎ)﷮2﷯ + 3 −1+ℎ﷯ − 5) − (−6)﷮h﷯ = lim﷮h→0﷯ 2(−1+ℎ)﷮2﷯ + 3 −1+ℎ﷯ − 5 + 6﷮h﷯ = lim﷮h→0﷯ 2 −1﷯2 + ℎ2 +2 −1﷯ℎ﷯ − 3 + 3ℎ + 1﷮h﷯ = lim﷮h→0﷯ 2 1 + ℎ2 − 2ℎ﷯ + 3ℎ − 2﷮h﷯ = lim﷮h→0﷯ 2 + 2ℎ2 − 4ℎ + 3ℎ − 2﷮h﷯ = lim﷮h→0﷯ 2ℎ2− ℎ﷮h﷯ = lim﷮h→0﷯ ℎ(2ℎ − 1)﷮h﷯ = lim﷮h→0﷯ 2h – 1 Putting h = 0 = 2(0) – 1 = – 1 Hence f’( –1) = – 1 Now, finding f’(0) For f’(0) f’(x)= lim﷮h→0﷯ 𝑓 𝑥 + ℎ﷯ − 𝑓(𝑥)﷮ℎ﷯ f’(x)= lim﷮h→0﷯ 2 𝑥 + ℎ﷯2 + 3 𝑥 + ℎ﷯ − 5﷯ −[2𝑥2 + 3𝑥 − 5]﷮ℎ﷯ putting x = 0 f’(0)= lim﷮h→0﷯ 2 0 + ℎ﷯2 + 3 0 + ℎ﷯ − 5﷯ −[2(0)2 + 3(0) − 5]﷮ℎ﷯ f’(0)= lim﷮h→0﷯ 2ℎ2 + 3ℎ − 5﷯ − [0 + 0 − 5]﷮ℎ﷯ f’(0)= lim﷮h→0﷯ 2ℎ2 + 3ℎ − 5 + 5﷮ℎ﷯ f’(0)= lim﷮h→0﷯ 2ℎ2 + 3ℎ﷮ℎ﷯ = lim﷮h→0﷯ ℎ(2ℎ+3)﷮h﷯ = lim﷮h→0﷯ 2h + 3 Putting h = 0 = 2(0) + 3 = 3 Hence, f’(0) = 3 Now, f’ (0) + 3f’( – 1) Putting value of f’(0) & f’( –1) = 3 + 3 ( –1) = 0 Hence Proved Example 6 (Method 2) Find the derivative of the function f(x) = 2x2 + 3x – 5 at x = –1. Also prove that f’(0) + 3f’( –1) = 0. Given f(x) = 2x2 + 3x – 5 Now, f’(x) = (2x2 + 3x – 5)’ = 2(2.x2–1) + 3(1.x1–1) – 0 = 2(2x1) + 3(1) = 4x + 3 Putting x = 0 f’(0) = 4(0) + 3 = 0 + 3 = 3 Taking f’ (0) + 3f’( – 1) Putting value of f’(0) & f’( –1) = 3 + 3 ( –1) = 3 – 3 = 0 Hence Proved