Find: e^{2x}(2x+1)=0 find the critical number?

Question:

{eq}f(x)=e^{2x}(2x+1) {/eq} find the critical number?

Critical Points of Functions:

It is to be remembered that the critical point of a function are those points at which the function has either a local maximum or a local minimum. This indirectly also means that critical points are those points at which the slope of the function should be 0 or the first derivative is 0. We can find the coordinate of critical points by differentiation and solving? the equation obtained by equating the first derivative to 0.

The critical point on a function is the point at which the first derivative of the function is 0.

{eq}\begin{align} f(x) &= e^{2x}(2x+1) \\ f'(x) &= 0 & \left[\text{ At critical point, first derivative is 0 } \right]\\ \Rightarrow \frac{d}{dx} \left[ e^{2x}(2x+1) \right] &= 0\\ \Rightarrow (2x+1) \frac{d}{dx} e^{2x} + e^{2x} \frac{d}{dx} (2x+1) &= 0\\ \Rightarrow (2x+1) (2e^{2x} ) + e^{2x} (2) &= 0\\ \Rightarrow 2e^{2x} (2x+1+1) &= 0\\ 2e^{2x} \neq 0 \Rightarrow (2x+2) &= 0 \; \\ \Rightarrow x+1 &= 0 \\ \Rightarrow x &= -1\\ \Rightarrow f(-1) &= e^{2(-1) }(2(-1) +1) \\ &= e^{-2 }(-2 +1) \\ &= -e^{-2 }\\ &= -0.1353 \end{align} {/eq}

Therefore, the coordinates of the critical point of the function are? {eq}\displaystyle \boxed{\color{blue} { (-1, -0.1353) }} {/eq}