elynnea4xl

2023-02-21

What are the inflections points of $y={e}^{2x}-{e}^{x}$?

Peruvianoe4p

Beginner2023-02-22Added 5 answers

By the Chain Rule, the first derivative is $y\prime =2{e}^{2x}-{e}^{x}$ and the second derivative is $y\prime \prime =4{e}^{2x}-{e}^{x}$. Inflection points occur at x values where the sign of the second derivative changes (from positive to negative or negative to positive).

Setting $y\prime \prime =0$ leads to $4{e}^{2x}-{e}^{x}=0$. The left-hand side of this equation can be factored as ${e}^{x}(4{e}^{x}-1)=0$. Since $e}^{x$ is never zero, it follows that we just need to solve $4{e}^{x}-1=0$ to get $x=\mathrm{ln}\left(\frac{1}{4}\right)=-\mathrm{ln}\left(4\right)\approx -1.386$.

You can check that $y\prime \prime =4{e}^{2x}-{e}^{x}$ changes sign at this point by graphing it. Therefore, there is an inflection point at $x=-\mathrm{ln}\left(4\right)$. The second coordinate of this point can be found by plugging it into the original function to get $e}^{2\cdot \mathrm{ln}\left(\frac{1}{4}\right)}-{e}^{\mathrm{ln}\left(\frac{1}{4}\right)}=\frac{1}{16}-\frac{1}{4}=-\frac{3}{16$. The inflection point is therefore $(-\mathrm{ln}\left(4\right),-\frac{3}{16})$.

Setting $y\prime \prime =0$ leads to $4{e}^{2x}-{e}^{x}=0$. The left-hand side of this equation can be factored as ${e}^{x}(4{e}^{x}-1)=0$. Since $e}^{x$ is never zero, it follows that we just need to solve $4{e}^{x}-1=0$ to get $x=\mathrm{ln}\left(\frac{1}{4}\right)=-\mathrm{ln}\left(4\right)\approx -1.386$.

You can check that $y\prime \prime =4{e}^{2x}-{e}^{x}$ changes sign at this point by graphing it. Therefore, there is an inflection point at $x=-\mathrm{ln}\left(4\right)$. The second coordinate of this point can be found by plugging it into the original function to get $e}^{2\cdot \mathrm{ln}\left(\frac{1}{4}\right)}-{e}^{\mathrm{ln}\left(\frac{1}{4}\right)}=\frac{1}{16}-\frac{1}{4}=-\frac{3}{16$. The inflection point is therefore $(-\mathrm{ln}\left(4\right),-\frac{3}{16})$.