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Question

Find the inflection points and intervals in which the function $$f(x)=\dfrac{x^{2}}{x^{2}+3}$$ is concave up and concave down.


Solution

$$f\left( x \right) = \dfrac{{{x^2}}}{{{x^2} + 3}}$$
$$f'\left( x \right) = \dfrac{{2x\left( {{x^2} + 3} \right) - 2{x^3}}}{{{{\left( {{x^2} + 3} \right)}^2}}}$$$$ = \dfrac{{ - 6x}}{{{{\left( {{x^2} + 3} \right)}^2}}}$$
$$  f''\left( x \right) = \dfrac{{6x{{\left( {{x^2} + 3} \right)}^2} - 6x \times 2\left( {{x^2} + 3} \right) \times 2x}}{{{{\left( {{x^2} + 3} \right)}^4}}}$$ 
$$ = \dfrac{{6x\left( {{x^2} + 3} \right)\left( {{x^2} - 4x + 3} \right)}}{{{{\left( {{x^2} + 3} \right)}^4}}}$$
$$ = \dfrac{{6x\left( {{x^2} + 3} \right)\left( {x - 1} \right)\left( {x - 3} \right)}}{{{{\left( {{x^2} + 3} \right)}^4}}}$$
For point of inflection
$$f''\left( x \right) = 0$$
$$ \Rightarrow \dfrac{{6x\left( {{x^2} + 3} \right)\left( {x - 1} \right)\left( {x - 3} \right)}}{{{{\left( {{x^2} + 3} \right)}^4}}} = 0$$
$$ \Rightarrow 6x\left( {{x^2} + 3} \right)\left( {x - 1} \right)\left( {x - 3} \right) = 0$$
$$ \Rightarrow x = 0,1,3$$
so, points of inflection are $$x = 0,1,3$$
for concave upward
$$f''\left( x \right) > 0$$
$$ \Rightarrow \dfrac{{6x\left( {{x^2} + 3} \right)\left( {x - 1} \right)\left( {x - 3} \right)}}{{{{\left( {{x^2} + 3} \right)}^4}}} > 0$$
$$ \Rightarrow 6x\left( {{x^2} + 3} \right)\left( {x - 1} \right)\left( {x - 3} \right) > 0$$
$$ \Rightarrow x \in \left( {0,1} \right) \cup \left( {3,\infty } \right)$$
For concave downward
$$f''\left( x \right) < 0$$
$$ \Rightarrow \dfrac{{6x\left( {{x^2} + 3} \right)\left( {x - 1} \right)\left( {x - 3} \right)}}{{{{\left( {{x^2} + 3} \right)}^4}}} < 0$$
$$ \Rightarrow 6x\left( {{x^2} + 3} \right)\left( {x - 1} \right)\left( {x - 3} \right) < 0$$
$$ \Rightarrow x \in \left( { - \infty ,0} \right) \cup \left( {1,3} \right)$$

Mathematics

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