Sketch the graph of the 2 ^{nd} Derivative, $\textcolor[rgb]{}{f}\textcolor[rgb]{}{\prime}\textcolor[rgb]{}{\prime}\textcolor[rgb]{}{\left(}\textcolor[rgb]{}{x}\textcolor[rgb]{}{\right)}$, determine the concavity of the original graph, $\textcolor[rgb]{}{f}\textcolor[rgb]{}{\left(}\textcolor[rgb]{}{x}\textcolor[rgb]{}{\right)}$, and identify any inflection points based on the given derivative graph.


Step 1: Identify and then boldly label the graph you are working with. In this example you have been given the derivative graph, the ${\textcolor[rgb]{}{f}}^{\textcolor[rgb]{}{\prime}}\textcolor[rgb]{}{\left(}\textcolor[rgb]{}{x}\textcolor[rgb]{}{\right)}$ graph. 


Step 2: Determine if you are being asked to go move down a level or move up a level. 
In this example you are being asked to move down a level. You are starting with the derivative graph, ${\textcolor[rgb]{}{f}}^{\textcolor[rgb]{}{\prime}}\textcolor[rgb]{}{\left(}\textcolor[rgb]{}{x}\textcolor[rgb]{}{\right)}$, and being asked to move down a level, find the 2 ^{nd} Derivativegraph, $\textcolor[rgb]{}{f}\textcolor[rgb]{}{\prime}\textcolor[rgb]{}{\prime}\textcolor[rgb]{}{\left(}\textcolor[rgb]{}{x}\textcolor[rgb]{}{\right)}$. 

Step 3 (If you are moving down ): Mark your given graph with plus (+), minus (), and zero (0) based on the 1 ^{st} Derivative Triangle. The 1 ^{st} Derivative of ${\textcolor[rgb]{}{f}}^{\textcolor[rgb]{}{\prime}}\textcolor[rgb]{}{\left(}\textcolor[rgb]{}{x}\textcolor[rgb]{}{\right)}$ is the 2 ^{nd} Derivative, $\textcolor[rgb]{}{f}\textcolor[rgb]{}{\prime}\textcolor[rgb]{}{\prime}\textcolor[rgb]{}{\left(}\textcolor[rgb]{}{x}\textcolor[rgb]{}{\right)}$.



Step 4 (If you are moving down ): Use your plus (+), minus (), and zero (0) that you marked along with their estimated slope to determine the yvalues on the graph you are moving down to.
The first points that you will always want to graph will be the zeros that you have marked. You know where a zero value will always be graphed, it will always be graphed on the xaxis . After that you want to work your way left to right keeping in mind that the steeper the graph the larger the number, and increasing means a positive slope, and decreasing means a negative slope. I find it helpful to use approximated values to keep track of it all. A steep positive slope I would use m=15 , or a flat negative slope I use m=1 in my head or writing on the actual graph. If you are given a piecewise graph with constants and linear equations , you can actually use real values if needed. The derivative of a constant is always zero, and the derivative of a linear equation ( y= m x+b ) is always the m , the slope.
$\left(\textcolor[rgb]{}{x},\textcolor[rgb]{}{y}\right)=\left(\textcolor[rgb]{}{x},\textcolor[rgb]{}{\mathit{slope}}\textcolor[rgb]{}{}\mathit{of\; the}\textcolor[rgb]{}{}\textcolor[rgb]{}{\mathit{Derivative\; Graph}}\right)$ $\left(\textcolor[rgb]{}{x},\textcolor[rgb]{}{y}\right)=\left(\textcolor[rgb]{}{x},\textcolor[rgb]{}{m}\right)$ 


Final Result: In this example you would draw a final version of your derivative graph based on the sketch from Step 4 . Remember to always make sure your final answer passes the vertical line test to ensure it is a function. The sketch of the 2 ^{nd} Derivative, $\textcolor[rgb]{}{f}\textcolor[rgb]{}{\prime}\textcolor[rgb]{}{\prime}\textcolor[rgb]{}{\left(}\textcolor[rgb]{}{x}\textcolor[rgb]{}{\right)}$, given this derivative graph, ${\textcolor[rgb]{}{f}}^{\textcolor[rgb]{}{\prime}}\textcolor[rgb]{}{\left(}\textcolor[rgb]{}{x}\textcolor[rgb]{}{\right)}$,would look like:
Concavity: Where the 2 ^{nd} Derivative graph has positive yvalues , the original graph, $\textcolor[rgb]{}{f}\textcolor[rgb]{}{\left(}\textcolor[rgb]{}{x}\textcolor[rgb]{}{\right)}$, is concave up. The xinterval would be $\left(\textcolor[rgb]{}{\u2013}\textcolor[rgb]{}{\mathrm{\infty}},\textcolor[rgb]{}{\u2013}\textcolor[rgb]{}{2}\right)\bigcup \left(\textcolor[rgb]{}{2},\textcolor[rgb]{}{\mathrm{\infty}}\right)$. Notice that you would not include the end points, and we use the union symbol to connect the two xintervals . Where the 2 ^{nd} Derivative graph has negative yvalues , the original graph, $\textcolor[rgb]{}{f}\textcolor[rgb]{}{\left(}\textcolor[rgb]{}{x}\textcolor[rgb]{}{\right)}$, is concave down. The xinterval would be $\left(\textcolor[rgb]{}{\u2013}\textcolor[rgb]{}{2},\textcolor[rgb]{}{2}\right)$.
Inflection Point: Inflection points occur when i) the 2 ^{nd} Derivative equals zero and ii) the concavity changes . Here you have two inflection points. First at x=2 and again at x=2 . At those two xvalues the 2 ^{nd} Derivative graph equals zero, the graph switches from positive to negative or negative to positive. 