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Exercise 7.10 · Q20

Q.The point of inflection of the curve y=(x−1)3y=(x-1)^3 is

(1) (0,0)(0,0)
(2) (0,1)(0,1)
(3) (1,0)(1,0)
(4) (1,1)(1,1)
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Differentiate twice, find where y′′=0y''=0, confirm the sign change, and evaluate yy there.

Step 1. Differentiate.

y=(x−1)3⇒y′=3(x−1)2⇒y′′=6(x−1)y=(x-1)^3\Rightarrow y'=3(x-1)^2\Rightarrow y''=6(x-1).

Step 2. Solve y′′=0y''=0.

6(x−1)=0⇒x=16(x-1)=0\Rightarrow x=1.

Step 3. Confirm the sign change. …

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