**Ex 5.8, 5**

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 5.8, 5 Verify Mean Value Theorem, if 𝑓 (𝑥) = 𝑥3 – 5𝑥2 – 3𝑥 in the interval [a, b], where a = 1 and b = 3. Find all 𝑐 ∈ (1, 3) for which 𝑓 ′(𝑐) = 0. 𝑓 (𝑥) = 𝑥3 – 5𝑥2 – 3𝑥 in [a, b], where a = 1 and b = 3 Condition 1 𝑓 (𝑥) = 𝑥3 – 5𝑥2 – 3𝑥 𝑓 𝑥 is a polynomial & Every polynomial function is continuous ⇒ 𝑓 𝑥 is continuous at 𝑥∈ 1, 3 Condition 2 𝑓 𝑥 = 𝑥3 – 5𝑥2 – 3𝑥 𝑓 𝑥 is a polynomial & Every polynomial function is differentiable ⇒ 𝑓 𝑥 is differentiable at 𝑥∈ 1, 3 Now 𝑓 𝑥 = 𝑥3 – 5𝑥2 – 3𝑥 𝑓′ 𝑥 = 3𝑥2 –10𝑥 – 3 𝑥∈ 1, 3 𝑓′ 𝑐 = 3 𝑐2 –10 𝑐 – 3 𝑥∈ 1, 3 𝑓′ 𝑐 = 3 𝑐2−10𝑐−3 Also, 𝑓 𝑥 = 𝑥3 – 5𝑥2 – 3𝑥 𝑓 𝑎 = 𝑓 1 = 13−5 12−3 1 = 1−5−3 = −7 𝑓 𝑏 = 𝑓 3 = 33−5 32−3 3 = 27−45−9 = −27 By Mean Value Theorem 𝑓′ 𝑐 = 𝑓 𝑏 − 𝑓 𝑎𝑏 − 𝑎 3 𝑐2−10𝑐−3 = −27 − −73 − 1 3 𝑐2−10𝑐−3 = −27 + 72 3 𝑐2−10𝑐−3 = −202 3 𝑐2−10𝑐−3 = −10 3 𝑐2−10𝑐−3+10 = 0 3 𝑐2−10𝑐+7 = 0 3 𝑐2−3𝑐−7𝑐+7 = 0 3𝑐 𝑐−1−7 𝑐−1 = 0 3𝑐−7 𝑐−1 = 0 So, c = 73 & c = 1 Now c = 73 lies between 1 & 3 c = 𝟕𝟑 ∈ 1, 3 Thus, Mean Value Theorem is verified. Now, We need to find c∈ 1, 3 For which 𝑓′ 𝑐 = 0 𝑓′ 𝑐 = 0 3 𝑐2−10𝑐−3 = 0 The above equation is of the form 𝐴 𝑥2+𝐵𝑥+𝐶 x = −𝐵 ± 𝐵2 −4𝐴𝐶 2𝐴 c = − −10 ± −102 − 4 −3 3 2𝐴 c = 10 ± 100 + 36 2 −3 c = 10 ± 136 6 c = 10 ± 2 × 2 × 346 c = 10 ± 2 346 c = 2 5 ± 34 6 c = 5 ± 343 So, Thus, c = 3.61 & c = –0.28 Both value does not lie between 1, 3 Hence, there exists no value of 𝐜∈ 1, 3 for which 𝑓′ 𝑐 = 0

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Davneet Singh

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