Question

Consider f(x)=x2x2+3f(x)=\dfrac{x^2}{x^2+3}. Find the local maximum and minimum values of ff.

Solution

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Using the solution for the exercise 10a, we have that f(x)=6x(x2+3)2f'(x)=\frac{6x}{\left(x^2+3\right)^2} and the critical point is

c=0,c=0,

which can be local maximum or local minimum. We calculate values of ff' for some negative and some positive point:

f(1)=6(1)((1)2+3)2=616<0,f(1)=61(12+3)2=616>0.\begin{align*} f(-1)&=\frac{6 \cdot (-1)}{\left((-1)^2+3\right)^2}=\frac{-6}{16}<0,\\ f(1)&=\frac{6 \cdot 1}{\left(1^2+3\right)^2}=\frac{6}{16}>0. \end{align*}

Since ff decreases at (,0)(-\infty,0) and then increases at (0,+)(0, +\infty), we get that $\text{\textcolor{#4257b2}{ f(0)=0f(0)=0 is local minimum. Function ff has no local maximum}}$.

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