Since the second derivative is negative, then the function at this point is convex downward.
Example 8.
Find the intervals of convexity/concavity of the function \[f\left( x \right) = \frac{1}{{1 + {x^2}}}.\]
Solution.
The function is defined and differentiable for all \(x \in \mathbb{R}.\) To determine the direction of convexity, we use the convexity test based on the second derivative. Calculate the second derivative:
Here in solving the inequalities, we have used the fact that the denominator in the expression for the second derivative is always positive: \({\left( {1 + {x^2}} \right)^3} \gt 0.\)
Thus, based on the sign of the second derivative, we establish that the given function is
If we try to solve the equation \(f^{\prime\prime}\left( x \right) = 0,\) we find that it has no solutions. But \(f^{\prime\prime}\left( x \right)\) is undefined at \(x = 0,\) so we plot this point on the sign chart to investigate the sign of the second derivative.
This tells us that the function is convex upward on \(\left( { -\infty, 0} \right)\) and convex downward on \(\left( {0, +\infty} \right).\)
Example 10.
Find the intervals of convexity/concavity of the function \[f\left( x \right) = x - \cos x.\]
Solution.
Differentiating successively, we calculate the second derivative:
Determine the intervals where the second derivative is positive and negative. The problem is reduced to the solution of the simple trigonometric inequality:
\[f^{\prime\prime}\left( x \right) \gt 0,\;\; \Rightarrow \cos x \gt 0,\;\; \Rightarrow x \in \left( { - \frac{\pi }{2} + 2\pi n,\frac{\pi }{2} + 2\pi n} \right),\;n \in \mathbb{Z};\]
\[f^{\prime\prime}\left( x \right) \lt 0,\;\; \Rightarrow \cos x \lt 0,\;\; \Rightarrow x \in \left( { \frac{\pi }{2} + 2\pi k,\frac{3\pi }{2} + 2\pi k} \right),\;k \in \mathbb{Z}.\]
Hence, the function is strictly convex downward for
\[x \in \left( - {\frac{\pi }{2}} + 2\pi n, {\frac{\pi }{2}} + 2\pi n \right), n \in \mathbb{Z},\]
and strictly convex upward for
\[x \in \left( {\frac{\pi }{2}} + 2\pi k, {\frac{3\pi }{2}} + 2\pi k \right), k \in \mathbb{Z}.\]
Note that the linear term \(x\) in the expression for the function does not affect the direction of convexity.
Example 11.
Find the intervals of convexity/concavity of the \(4\)th order polynomial function:
\[f\left( x \right) = {x^4} + 2{x^3} - 36{x^2} + 2x + 1.\]
Solution.
Differentiating successively, we find the second derivative:
We solve the equation \(f^{\prime\prime}\left( x \right) = 0\) and determine the sign of the second derivative in the corresponding intervals (Figure \(7\)):
Consequently, this function (as well as its curve) is strictly convex downward in the intervals \(\left( { - \infty , - 3} \right)\) and \(\left( {2, + \infty } \right)\) and is strictly convex upward in the interval \(\left( {-3, 2} \right).\)
Example 12.
Determine where the function \[f\left( x \right) = {x^4} - 2{x^3} - 12{x^2} + 1\] is convex upward and convex downward?
Solution.
We use the Convexity/Concavity Test. Find the second derivative:
It follows from the sign chart that \(f\left( x \right)\) is convex downward on the intervals \(\left( { - \infty , - 1} \right)\) and \(\left( {2, + \infty } \right)\) and convex upward on the interval \(\left( { - 1, 2} \right).\)
Example 13.
Find the intervals on which the function \[f\left( x \right) = {x^4} - 6{x^2} - 6x + 1\] is convex upward and convex downward.
Hence, the function is convex downward on \(\left( { - \infty , - 1} \right)\) and \(\left( { 1, +\infty} \right)\) and convex upward on \(\left( { -1, - 1} \right).\)
Example 14.
Find the intervals of convexity/concavity of the function \[f\left( x \right) = \frac{{{x^3}}}{{1 + {x^2}}}.\]
Solution.
Obviously, this function is defined and differentiable for all \(x.\) Find successively the first and second derivative:
It follows from Figure \(10\) that the second derivative is positive in the intervals \(\left( { - \infty , - \sqrt 3 } \right)\) and \(\left( {0, \sqrt 3 } \right).\) Hence, the function is strictly convex downward here. In the other two intervals − at \(x \in \left( {-\sqrt 3, 0} \right)\) and at \(x \in \left( {\sqrt 3, +\infty} \right)\) − the function is strictly convex upward as shown schematically in Figure \(10.\)
Example 15.
Determine the intervals on which the function \[f\left( x \right) = {e^{\sqrt x }}\] is convex upward and convex downward.
Solution.
Find \(f^\prime\) and \(f^{\prime\prime}:\)
\[f^\prime\left( x \right) = \left( {{e^{\sqrt x }}} \right)^\prime = {e^{\sqrt x }} \cdot \left( {\sqrt x } \right)^\prime = \frac{{{e^{\sqrt x }}}}{{2\sqrt x }};\]
\[f^{\prime\prime}\left( x \right) = \left( {\frac{{{e^{\sqrt x }}}}{{2\sqrt x }}} \right)^\prime = \frac{{\left( {{e^{\sqrt x }}} \right)^\prime \cdot \sqrt x - {e^{\sqrt x }} \cdot \left( {\sqrt x } \right)^\prime}}{{{{\left( {\sqrt x } \right)}^2}}} = \frac{{\frac{{{e^{\sqrt x }}}}{{2\sqrt x }} \cdot \sqrt x - {e^{\sqrt x }} \cdot \frac{1}{{2\sqrt x }}}}{x} = \frac{{\frac{{{e^{\sqrt x }}}}{{2\sqrt x }}\left( {\sqrt x - 1} \right)}}{x} = \frac{{{e^{\sqrt x }}\left( {\sqrt x - 1} \right)}}{{2\sqrt {{x^3}} }}.\]
We see that \(f^{\prime\prime}\left( x \right)\) does not exist at \(x = 0\) and it is equal to zero at \(x = 1\) (which is a point of inflection).
As it follows from the sign chart (see above), the function is convex downward on \(\left( {1, + \infty } \right)\) and convex upward on \(\left( {0, 1} \right).\)
Example 16.
Find the intervals on which the function \[f\left( x \right) = {e^{ - {x^3}}}\] is convex upward and convex downward.
As you can see from the sign chart, the function is convex downward on the intervals \(\left( { - \infty ,0} \right)\) and \(\left( { \sqrt[3]{{\frac{2}{3}}}, +\infty} \right).\) Respectively, the function is convex upward on \(\left( { 0, \sqrt[3]{{\frac{2}{3}}}} \right).\)
The graph of the function is schematically given above.
Example 17.
Find the intervals on which the function \[f\left( x \right) = {e^{\frac{1}{x}}}\] is convex upward and convex downward.
Solve the equation \(f^{\prime\prime}\left( x \right) = 0\) and draw a sign chart for \(f^{\prime\prime}\left( x \right).\)
\[f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow \frac{{{e^{\frac{1}{x}}}\left( {1 + 2x} \right)}}{{{x^4}}} = 0,\;\; \Rightarrow x = - \frac{1}{2}.\]
Note that the function and its derivatives do not exist at \(x = 0,\) so we also indicate this point on the sign chart.
Thus, the function is convex downward on \(\left( {-\frac{1}{2}, 0} \right)\) and \(\left( { 0, +\infty} \right)\) and convex upward on \(\left( { -\infty, -\frac{1}{2}} \right).\)
Example 18.
Investigate the direction of convexity of the curve defined by the parametric equations
\[x = a\left( {t - \sin t} \right), y = a\left( {1 - \cos t} \right),\]
where \(a \gt 0.\)
Solution.
This curve is a cycloid. Calculate the first derivative:
As seen, the derivative \(y^{\prime\prime}\) is negative everywhere except at the points \(t = 2\pi n,\) \(n \in \mathbb{Z}.\) For these points, the \(x\)-coordinate is
Since the exponential function is always positive, the second derivative is also positive for all \(x \in \mathbb{R}.\) Consequently, the given curve is everywhere strictly convex downward.