Concavity
Concavity pertains to the shape of a function and how it changes. This change relates to its rate of change (also known as slope). Concavity comes in 2 forms:
- Concave Up (or CU) occurs when the slope of \(f(x)\) is increasing
- Concave Down (or CU) occurs when the slope of \(f(x)\) is decreasing
Graphically, we can determine that the function is CU since it's always above the tangent line. Conversely, we can determine that the function is CD since it's always below the tangent line. Likewise, a function is CU when it opens upward and CD when it opens downward.
In order to determine the concavity of a function, we can find its second derivative. We can then conduct the Second Derivative Test. This involves solving \(f''(x) = 0\) or \(f''(x) = \text{Undefined}\) by finding test \(x\)-values which will result in a potential Point of Inflection. The test value can be picked from any value within an interval between critical points or end points.
The Second Derivative Test states:
- If \(f''(x) > 0\), this means \(f(x)\) is Concave Up
- If \(f''(x) < 0\), this means \(f(x)\) is Concave Down
We can then identify if a point is a Point of Inflection if the concavity changes between intervals from positive to negative or vice versa.
NOTE: A function can still increase and be considered Concave Down and vice versa. The most important aspect of determining the concavity depends on the slope of the original function!
Example
- Analyze any Points of Inflection for \(f(x) = \cfrac{1}{3}x^3 - \cfrac{5}{2}x^2 + 6x-1\)
- Graph \(f(x)\) and \(f''(x)\) and explain how they indicate concavity
First, can find the First Derivative:
\(f'(x) = \cancel{3} \left(\cfrac{1}{\cancel{3}}x^2\right) - \cancel{2} \left(\cfrac{5}{\cancel{2}}x^2\right) + 6x\)
\(f'(x) = x^2 - 5x + 6\)
Review Intervals of Change for how to use the First Derivative to find the characteristics of a function.
Next, in order to determine the Point of Inflection, we can find the second derivative:
\(f''(x) = 2x - 5\)
Then, we can set the equation to \(0\) to determine the Point of Inflection:
\(0 = 2x - 5\)
\(2x = 5\)
Finally, we can conduct the Second Derivatives Test to determine what occurs around the Point of Inflection:
| Interval | Test \(x\)-value | \(f''(x)\) | Conclusion |
|---|---|---|---|
| \(\textcolor{red}{(-∞,5/2)}\) | \(1\) | \(f''(2)= -1\) | \(f\) is Concave Down |
| \(\textcolor{green}{(5/2, ∞)}\) | \(3\) | \(f''(3)= 1\) | \(f\) is Concave Up |
Since the function changes concavity between intervals, we can determine that \(x = \cfrac{5}{2}\) represents a Point of Inflection.
We can determine the interval the function is Concave Up as \(\textcolor{green}{(5/2, ∞)}\).
We can also determine the interval the function is Concave Down as \(\textcolor{red}{(-∞,5/2)}\).
We can graph \(f(x)\) as such:
From this graph, we can determine:
- \(f(x)\) opening downward means that \(f(x)\) is Concave Down
- \(f(x)\) opening upward means that \(f(x)\) is Concave Up
We can graph \(f''(x)\) as such:
From this graph, we can determine:
- \(f''(x)\) below the \(x\)-axis means that \(f(x)\) is Concave Down
- \(f''(x)\) above the \(x\)-axis means that \(f(x)\) is Concave Up
First, we can find the First Derivative. We can do so by using Quotient Rule:
\(h'(x) = \cfrac{\textcolor{blue}{g(x)}\textcolor{red}{f'(x)} - \textcolor{red}{f(x)}\textcolor{blue}{g'(x)}}{[\textcolor{blue}{g(x)}]^2}\)
\(h'(x) = \cfrac{(\textcolor{blue}{x^2 + 4})(\textcolor{red}{0}) - (\textcolor{red}{2})(\textcolor{blue}{2x})}{[\textcolor{blue}{x^2 + 4}]^2}\)
\(h'(x) = \cfrac{-4x}{(x^2 + 4)^2}\)
Next, in order to determine the Point of Inflection, we can find the second derivative. Like before, we can use Quotient Rule:
\(h''(x) = \cfrac{\textcolor{blue}{g(x)}\textcolor{red}{f'(x)} - \textcolor{red}{f(x)}\textcolor{blue}{g'(x)}}{[\textcolor{blue}{g(x)}]^2}\)
\(h''(x) = \cfrac{(\textcolor{blue}{(x^2 + 4)^2})(\textcolor{red}{-4}) - (\textcolor{red}{-4x})(\textcolor{blue}{2(x^2 + 4)(2x)})}{[(x^2 + 4)^2]^2}\)
\(h''(x) = \cfrac{-4(x^2 + 4)^2 + 16x^2(x^2 + 4)}{(x^2 + 4)^4}\)
Then, we can simplify the expression by taking a common factor out of the numerator:
\(h''(x) = \cfrac{\cancel{(x^2+4)}[-4(x^2 + 4) + 16x^2]}{(x^2 + 4)^{\cancel{4}}}\)
\(h''(x) = \cfrac{-4(x^2 + 4) + 16x^2}{(x^2 + 4)^3}\)
We can simplify further by expanding the numerator and collecting like terms:
\(h''(x) = \cfrac{-4x^2 - 16 + 16x^2}{(x^2 + 4)^3}\)
\(h''(x) = \cfrac{12x^2 - 16}{(x^2 + 4)^3}\)
\(h''(x) = \cfrac{4(3x^2 - 4)}{(x^2 + 4)^3}\)
\(h''(x) = \cfrac{4(\sqrt{3}x + 2)(\sqrt{3}x - 2)}{(x^2 + 4)^3}\)
After, we can set the numerator equal to \(0\) to determine our possible Points of Inflection.
We can determine the first Point of Inflection as such:
\(0 = \sqrt{3}x + 2\)
\(\cfrac{\cancel{\sqrt{3}}x}{\cancel{\sqrt{3}}} = \cfrac{-2}{\sqrt{3}}\)
\(x_1 = -\cfrac{2}{\sqrt{3}}\)
We can determine the second Point of Inflection as such:
\(0 = \sqrt{3}x - 2\)
\(\cfrac{\cancel{\sqrt{3}}x}{\cancel{\sqrt{3}}} = \cfrac{2}{\sqrt{3}}\)
\(x_2 = \cfrac{2}{\sqrt{3}}\)
We can determine that our possible Points of Inflection are \(x = -\cfrac{2}{\sqrt{3}}, \cfrac{2}{\sqrt{3}}\).
Finally, we can conduct the Second Derivatives Test to determine what occurs around the Point of Inflection:
| Interval | Test \(x\)-value | \(h''(x)\) | Conclusion |
|---|---|---|---|
| \(\textcolor{green}{(-∞,-2/\sqrt{3})}\) | \(-\sqrt{3}\) | \(h''(-\sqrt{3})= 20/343\) | \(h\) is Concave Up |
| \(\textcolor{red}{(-2/\sqrt{3}, 2/\sqrt{3})}\) | \(0\) | \(h''(0)= -16/64\) | \(h\) is Concave Down |
| \(\textcolor{green}{(2/\sqrt{3}, ∞)}\) | \(\sqrt{3}\) | \(h''(\sqrt{3})= 20/343\) | \(h\) is Concave Up |
We can determine that both points are actual Points of Inflection.
Now that we have identified the Point of Inflection, we can now sketch a graph:
