SOLUTION:
Case: Inflexion points
A point of inflection is found where the graph (or image) of a function changes concavity. To find this algebraically, we want to find where the second derivative of the function changes sign, from negative to positive, or vice-versa. So, we find the second derivative of the given function.
Given:
[tex]f\mleft(x\mright)\text{ = }e^{-8x^2}[/tex]
Required:
a) Find the number of critical points
b) Find the x-coordinates of the critical points
Method:
Steps1:
[tex]\begin{gathered} f\mleft(x\mright)=\text{ }e^{-8x^2} \\ y=\text{e}^{-8x^2} \\ using\text{ chain rule} \\ \frac{dy}{dx}=\text{ 2}\times-8x^2\text{e}^{-8x^2} \\ \frac{dy}{dx}=-16x^2(\text{e})^{-8x^2} \end{gathered}[/tex]
We find the second derivative
[tex]\begin{gathered} \frac{dy}{dx}=-16x^2\text{e}^{-8x^2} \\ \frac{d^2y}{dx^2}=256x^2\text{e}^{-8x^2}\text{ - 16}e^{-8x^2} \\ \frac{d^{2}y}{dx^{2}}=\text{ 16}e^{-8x^2}\left(16x^2-1\right? \\ At\text{ critical points, the second derivative equals zero} \\ \text{16}e^{-8x^2}\left(16x^2-1\right?=\text{ 0} \\ \text{16}e^{-8x^2}\text{= 0 OR 16x}^2-1=0 \\ x\text{ is undefined OR 16x}^2=\text{ 1} \\ x^2=\frac{1}{16} \\ x=\text{ }\sqrt{\frac{1}{16}} \\ x=\text{ +}\frac{1}{4}\text{ or -}\frac{1}{4} \end{gathered}[/tex]
Final answer:
a) The function has 2 inflexion points
b) the values of x are -1/4 and +1/4
