Question

How do you find the points of inflection of the curve \[y = {e^{{x^2}}}\]?

Answer 1

Hint: We can find the points of inflection by differentiating twice, setting the result to zero, and checking whether the result is a genuine point of inflexion. If y” > 0, then the function is concave up and if y” < 0, then the function is concave down.

Complete step-by-step solution:
Let us write the given curve as
\[\Rightarrow y = {e^{{x^2}}}\] …………………… 1
To find the points of inflection we need to find the derivative, hence compute the first derivative of the given curve.
Hence, by equation 1 we get
\[\Rightarrow y' = 2x{e^{{x^2}}}\]
To find the second derivative let us apply the product rule as
\[\Rightarrow y'' = 2\left( {{e^{{x^2}}}} \right) + 2x\left( {2x} \right){e^{{x^2}}}\]
\[\Rightarrow y'' = 2{e^{{x^2}}} + 4{x^2}{e^{{x^2}}}\] ………………………. 2
Set equation 2 to zero to determine inflection points
\[\Rightarrow 2{e^{{x^2}}} + 4{x^2}{e^{{x^2}}} = 0\]
Simplifying the terms, we get
\[\Rightarrow 2{e^{{x^2}}}\left( {1 + 2{x^2}} \right) = 0\] ………………………. 3
We see this has no solution because \[2{e^{{x^2}}} \ne 0\] for all values of x. Furthermore, the third equation states that
\[\Rightarrow 1 + 2{x^2} = 0\]
\[\Rightarrow 2{x^2} = - 1\]
\[ \Rightarrow x = \sqrt { - \dfrac{1}{2}} \]
Hence, the value of x is
\[x = \sqrt { - \dfrac{1}{2}} \]
This doesn't have a real value so no real solution to this equation. This simply means the function \[y = {e^{{x^2}}}\]will have no inflection points and \[y''\]is positive on all its domain, therefore concave up on
\[\left( { - \infty ,\infty } \right)\].

Additional information:
Inflection point is a point on the graph of a function at which the concavity changes. Points of inflection can occur where the second derivative is zero. It means that the function changes from concave down to concave up or vice versa. In other words, the point in which the rate of change of slope from increasing to decreasing manner or vice versa is known as an inflection point. Those points are certainly not local maxima or minima.

Note: We must not assume that any point where \[y'' = 0\] is an inflection point. Instead, we need to see if the second derivative changes signs at those points and the function is defined at those points. If y” > 0, then the function is concave up and if y” < 0, then the function is concave down.