Economics at Zero Marginal Cost: Finding the Point of Inflection of a Function

Find the point of inflection of the function $\\f(x)=2x^{3}+6x^{2}-5x+1\\$

Solution:

Let $\\y=f(x)=2x^{3}+6x^{2}-5x+1\\$

To find: The points of inflection of the function

Differentiating the function twice to get the second derivative:

First derivative:

$\\\\\frac{dy}{dx}=\frac{d}{dx} (2x^{3}+6x^{2}-5x+1)\\ \frac{dy}{dx}=\frac{d}{dx}2x^{3}+\frac{d}{dx}6x^{2}-\frac{d}{dx}5x+\frac{d}{dx}1\\ \frac{dy}{dx}=2\frac{d}{dx}x^{3}+6\frac{d}{dx}x^{2}-5\frac{d}{dx}x+0\\ \frac{dy}{dx}=2.3x^{3-1}+6.2x^{2-1}-5.1\\ \frac{dy}{dx}=6x^{2}+12x-5\\$

Second derivative:

$\\\\\frac{d^{2}y}{dx^{2}}=\frac{d}{dx} (6x^{2}+12x-5)\\ \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}6x^{2}+\frac{d}{dx}12x-\frac{d}{dx}5\\ \frac{d^{2}y}{dx^{2}}=6\frac{d}{dx}x^{2}+12\frac{d}{dx}x-0\\ \frac{d^{2}y}{dx^{2}}=6.2x^{2-1}+12.1\\ \frac{d^{2}y}{dx^{2}}=12x+12\\$

First-order condition (Necessary condition):

$\\\\\frac{d^{2}y}{dx^{2}}=0\ or\ 12x+12=0\\$

Solving for x,

$\\\\12x+12=0\\ \12x=-12\\ \\x=\frac{-12}{12}\\ \\x=-1$

∴ Stationary point is at x=−1.

Second-order condition (Sufficient condition):

Immediate neighbour of x=−1 are x=−2 and x=0

$\\\\At\ x=-2\\\\ \frac{d^{2}y}{dx^{2}}=12x+12=12(-2)+12=-24+12=-12<0\ \\\\At\ x=0\\\\ \frac{d^{2}y}{dx^{2}}=12x+12=12(0)+12=0+12=12>0\\$

∴The function $\\f(x)=2x^{3}+6x^{2}-5x+1\\$ is concave down( $\frac{d^{2}y}{dx^{2}}<0\$ ) for x< −1 and it is concave up ( $\frac{d^{2}y}{dx^{2}}>0\$ ) for x>−1.

Hence, the graph of the function changes its curvature and x=−1 is an inflection point.

Now, the corresponding y−coordinate is

$\\\\y=f(x)=2x^{3}+6x^{2}-5x+1\\ \\y=f(-1)=2(-1)^{3}+6(-1)^{2}-5(-1)+1\\ \\y=f(-1)=2(-1)+6(1)+5+1\\ \\y=f(-1)=-2+6+6\\ \\y=f(-1)=-2+12\\ \\y=f(-1)=10\\$