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NSW Y12 Maths - Advanced Differential Calculus Concavity and Points of Inflexion

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Concavity and Points of Inflexion Theory

\textbf{Concavity:}\\ If \(f^{\prime \prime}(x)>0\) the curve is concave upwards.\\ If \(f^{\prime \prime}(x)<0\), the curve is concave downwards.\\ \textbf{Points of inflexion}\\ If \(f^{\prime \prime}(x)=0\), and concavity changes, it is a point of inflexion.\\ If \(f^{\prime}(x)=0\) also, it is a horizontal point of inflexion.\\ \begin{multicols}{2} \textbf{Example 1}\\ %51959 For the curve \(y=x^3-6x^2+5x-2\), determine the values of \(x\) for which the curve is concave up.\\ \textbf{Example 1 solution}\\ \(\begin{aligned} y&=x^3-6 x^2+5 x-2 \\ y^{\prime}&=3 x^2-12 x+5 \\ y^{\prime \prime}&=6 x-12 \end{aligned}\)\\ \(\begin{aligned} \text { Concave up} &\rightarrow y^{\prime \prime}>0 \end{aligned}\)\\ \(\begin{aligned} \therefore 6 x-12&>0 \\ x&>2 \end{aligned}\)\\ \columnbreak \textbf{Example 2}\\%10274 The points of inflection of the curve \(y = {x^4} - 6{x^2} + 2x - 4\) are?\\ \textbf{Example 2 solution}\\ \(\begin{aligned} y&=x^4-6x^2+2x-4\\ y'&=4x^3-12x+2\\ y''&=12x^2-12 \end{aligned}\)\\ \(\begin{aligned} \text{Point of inflection occur}&\text{when } y''=0 \end{aligned}\)\\ \(\begin{aligned} 12x^2-12&=0\\ x^2-1&=0\\ (x-1)(x+1)&=0\\ x=1 \text{ or } x&=-1 \end{aligned}\)\\ \(\begin{aligned} \text{At } x=1,\quad y=1-6+&2-4=-7 \rightarrow (1,-7)\\ \text{At } x=-1,\, y=1-6-&2-4=-11 \rightarrow (-1,-11) \end{aligned}\)\\ \(\text{Testing }(1,-7)\)\\ \(\begin{array}{r} \text{At }x=0.9 ,\;\; y''<0\\ \text{At }x=1.1 ,\;\; y''>0\\ \end{array}\)\(\begin{array}{r} \text{Concavity change }\end{array}\)\\ \(\therefore \text{ point of inflection}\)\\ \(\text{Testing }(-1,-11)\)\\ \(\begin{array}{r}\text{At }x=-1.1 ,\;\; y''>0\\\text{At }x=-0.9 ,\;\; y''<0\\\end{array} \begin{array}{r} \text{Concavity change } \end{array}\)\\ \(\therefore \text{ point of inflection}\) \end{multicols}

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