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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