The point of inflection for the function f(x)=sin−1x is:
A
x=0
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B
x=12
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C
x=−12
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D
x=1√2
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Solution
The correct option is Ax=0 Given: f(x)=sin−1x ⇒f′(x)=1√1−x2=(1−x2)−1/2 ⇒f′′(x)=−2x(−12)(1−x2)−3/2 ⇒f′′(x)=x(1−x2)32
Clearly, the second derivative is zero at x=0 and also it changes sign at x=0. Hence it is a point of inflection.