    Question

# For the function y=11+x2, which of the following is/are correct:

A
It has a horizontal asymptote.
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B
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C
It is increasing xR
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D
It has maxima at x=0
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Solution

## The correct option is D It has maxima at x=0The function is defined for all real values of x, Consequently, it has no vertical asymptotes. Since limx→±∞y(x)=limx→±∞11+x2=0, then the graph of the function has horizontal asymptote y=0, that is the x−axis is the horizontal asymptote. This function is even. y(–x)=11+(–x)2=11+x2=y(x). It is obvious that the function has no roots and positive for all x. At the point x=0, its value is y(0)=11+02=1 The first derivative: y′(x)=(11+x2)′ =–1(1+x2)2⋅(1+x2)′=–2x(1+x2)2. This shows that x=0 is a stationary point. When passing through this point the derivative changes sign from plus to minus. Therefore, we have a maximum at x=0. Its value is y(0)=1. Calculate the second derivative: y′′(x)=(–2x(1+x2)2)′ =–2(1+x2)2–2⋅2x⋅2x(1+x2)(1+x2)4 =8x2–2–2x2(1+x2)3=6x2–2(1+x2)3. It is equal to zero at the following points: y′′(x)=0⇒6x2–2(1+x2)3=0 ⇒2(√3x–1)(√3x+1)(1+x2)3=0,⇒x1=–1√3,x2=1√3. When passing through these points the second derivative changes its sign. Therefore, both points are inflection points. The function is strictly convex downward in the intervals (−∞,−1√3) and (1√3,+∞) and accordingly, strictly convex upward in the interval (−1√3,1√3) Since the function is even, the found inflection points have the same values of y: y(±1√3)=11+(±1√3)2=11+13=34. Figure presents a schematic graph of the function Hence, we can conclude clearly from the graph that it has a horizontal asymtote and it is symmetrical about x=0 or the y axis. Also it has maxima at x=0  Suggest Corrections  0      Similar questions  Explore more