A concave down function is a function where no line segment that joins 2 points on its graph ever goes above the graph. Intuitively, the graph is shaped like a hill. A concave up function, on the other hand, is a function where no line segment that joins 2 points on its graph ever goes below the graph. It is shaped like a U. In the graph above, the red curve is concave up, while the green curve is concave down. Functions in general have both concave up and concave down intervals. Inflection points exist when a function changes concavity.
A function can also have more than 1 root.
It’s easy to see this point on a graph.
Say you need to find the inflection point of the function below. f(x)=x3+2x−1{\displaystyle f(x)=x^{3}+2x-1} Use the power rule. f′(x)=3x3−1+2x1−1=3x2+2{\displaystyle {\begin{aligned}f^{\prime }(x)&=3x^{3-1}+2x^{1-1}\&=3x^{2}+2\end{aligned}}}
f′′(x)=(2)(3)x2−1=6x{\displaystyle {\begin{aligned}f^{\prime \prime }(x)&=(2)(3)x^{2-1}\&=6x\end{aligned}}}
6x=0x=0{\displaystyle {\begin{aligned}6x&=0\x&=0\end{aligned}}}
Remember that you are looking for sign changes, not evaluating the value. In more complicated expressions, substitution may be undesirable, but careful attention to signs often nets the answer much more quickly. For example, instead of evaluating numbers immediately, we could instead look at certain terms and judge them to be positive or negative. In our example, f′′(x)=6x. {\displaystyle f^{\prime \prime }(x)=6x. } Then plugging a negative x{\displaystyle x} yields a negative f′′(x),{\displaystyle f^{\prime \prime }(x),} while plugging a positive x{\displaystyle x} yields a positive f′′(x). {\displaystyle f^{\prime \prime }(x). } Therefore, x=0{\displaystyle x=0} is an inflection point of the function f(x)=x3+2x−1. {\displaystyle f(x)=x^{3}+2x-1. } There was no need to actually evaluate for our chosen values.
f(0)=(0)3+2(0)−1=−1. {\displaystyle f(0)=(0)^{3}+2(0)-1=-1. }
For example, if you do get an answer where x=0,{\displaystyle x=0,} you would test the subintervals by graphing (−infinity,0){\displaystyle (-infinity,0)} and (0,infinity){\displaystyle (0,infinity)}. Therefore, the inflection point is at 0.
For example, if you were given the task of finding whether or not h(x)=x2+4x{\displaystyle h(x)=x^{2}+4x} has an inflection point, you’d consider h′′{\displaystyle h^{\prime \prime }}, NOT h′{\displaystyle h^{\prime }}. This is because h′′{\displaystyle h^{\prime \prime }} is the second derivative, while h′{\displaystyle h^{\prime }} is the relative minimum point (which you aren’t looking for here).
For example, if your possible inflection points are x=−1{\displaystyle x=-1} and x=7,{\displaystyle x=7,} you would test the x values at (−infinity,−1),(−1,7),{\displaystyle (-infinity,-1),(-1,7),} and (7,infinity). {\displaystyle (7,infinity). } This would tell you that your second derivative has inflection points at both x=−1{\displaystyle x=-1} AND x=7. {\displaystyle x=7. }
This is true on both the TI-84 and the TI-89, but it may not be the exact same on older models.
For example, the function might be y1=x3−9/2x2−12x+3{\displaystyle y1=x^{3}-9/2x^{2}-12x+3}
Don’t worry if your screen doesn’t show the whole graph just yet—you’ll be able to adjust it.
You might have to go back and adjust this a few times, as it can be hard to figure out where your graph is exactly.
This is—you guessed it—how to tell your calculator to calculate inflection points.
This is how you’ll get your calculator to make a guess as to where the inflection point is. Now you have your answer!
