Course:MATH102/Question Challenge/2008 December Q1.5

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Question

The function $y=f(x)=\arctan(x)-(x/2)$ has local maXima (LX), local minima (LM) and inflection points(IP) as follows:

(a) LX: $x=1$, LM: $x=-1$, IP: $x=0$.

(b) LX: $x=-1$, LM: $x=1$, IP: $x=0$.

(c) LX: $x=-1$, LM: $x=1$, IP: none

(d) LX: $x={\sqrt {3}}$, LM: $x=-{\sqrt {3}}$, IP: $x=0$.

(e) LX: $x=-{\sqrt {3}}$, LM: $x={\sqrt {3}}$, IP: $x=0$.

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Solutions

Solution

$f(x)=\arctan(x)-{\frac {x}{2}}$ We take the first derivative to find the critical points: $f'(x)={\frac {1}{1+x^{2}}}-{\frac {1}{2}}$ $0={\frac {1}{1+x^{2}}}-{\frac {1}{2}}$ ${\frac {1}{2}}={\frac {1}{1+x^{2}}}$ $1+x^{2}=2$ $x^{2}=1$ $x=\pm 1$ We take the second derivative to find which critical points are maxima and which are minima: $f''(x)=-{\frac {2x}{(x^{2}+1)^{2}}}$ From this we already see that $f''(x)<0$ whenever $x>0$, and $f''(x)>0$ whenever $x<0$. This already tells us that the positive critical point is a local maximum, and the negative critical point is a local minimum. (It is optional to plug in the actual values, as only the sign of $f''(x)$ actually matters, but if we do plug in, we would get the following:) At $x=1$: $f''(1)=-2(1)/((1)^{2}+1)^{2})=-1/2<0$ At $x=-1$ $f''(-1)=-2(-1)/((-1)^{2}+1)^{2})=1/2>0$ The function $f(x)$ has a maximum at $x=1$ and a minimum at $x=-1$ We ask where the second derivative changes sign (and hence where it also equals 0) to find the inflection points: $0=-{\frac {2x}{(x^{2}+1)^{2}}}$ $0=-2x$ $x=0$. We observe that indeed the second derivative changes sign at $x=0$. Hence, the function$f(x)$ has an inflection point at $x=0$ LX: x = 1, LM: x = -1, IP: x = 0 The correct answer is A)