Question 3 If 𝑓 : [– 5, 5] → 𝐑 is a differentiable function and if 𝑓 ′(𝑥) does not vanish anywhere, then prove that 𝑓 (–5) ≠ 𝑓 (5). 𝑓 : [– 5, 5] → 𝐑 is a differentiable ⇒ We know that every differentiable function is continuous. Therefore f is continuous & differentiable both on (−5, 5) By Mean Value Theorem There exist some c in (5, −5) Such that 𝑓^′ (𝑐)=(𝑓(𝑏) − 𝑓(𝑎))/(𝑏 − 𝑎) Given that 𝑓^′ (𝑥) does not vanish any where ⇒ 𝑓^′ (𝑥) ≠ 0 for any value of x Thus, 𝑓^′ (𝑐) ≠ 0 (𝑓(5) − 𝑓(−5))/(5 −(−5) ) ≠ 0 (𝑓(5) − 𝑓(−5))/(5 + 5) ≠ 0 𝑓(5)− 𝑓(−5) ≠ 0 × 10 𝑓(5)− 𝑓(−5) ≠ 0 𝑓(5) "≠" 𝑓(−5) Hence proved.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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