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Exercise 7.3 · Q3

Q.Explain why Lagrange's mean value theorem is not applicable to the following functions in the respective intervals:

(i) f(x)=x+1x, x∈[−1,2]f(x)=\dfrac{x+1}{x},\ x\in[-1,2]
(ii) f(x)=∣3x+1∣, x∈[−1,3]f(x)=|3x+1|,\ x\in[-1,3]
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✓ Free question

LMVT needs continuity on [a,b][a,b] and differentiability on (a,b)(a,b); each part violates exactly one.

Step 1 (i). f(x)=x+1x, x∈[−1,2]f(x)=\dfrac{x+1}{x},\ x\in[-1,2].

ff is undefined at x=0x=0 (division by zero), and 0∈[−1,2]0\in[-1,2]. So ff is not continuous on [−1,2][-1,2] — LMVT does not apply.

Step 2 (ii). f(x)=∣3x+1∣, x∈[−1,3]f(x)=|3x+1|,\ x\in[-1,3].

ff is continuous everywhere (absolute value of a continuous function), so continuity on [−1,3][-1,3] holds. But ff has a corner (non-differentiable point) where 3x+1=03x+1=0, i.e. at x=−13x=-\dfrac13, and −13∈(−1,3)-\dfrac13\in(-1,3). So ff is not differentiable on the whole open interval (−1,3)(-1,3) — LMVT does not apply.

✓Final answer

(i) ff is undefined at x=0x=0, an interior point of [−1,2][-1,2], so continuity fails. (ii) ff has a non-differentiable corner at x=−13∈(−1,3)x=-\tfrac13\in(-1,3), so differentiability fails (even though ff is continuous throughout).

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