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.

Answer and Explanation:

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}


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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
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