Consider the function f(x)=xe^2x. Find the intervals on which f(x) is concave up and the...


Consider the function {eq}f(x)=xe^{2x} {/eq}. Find the intervals on which f(x) is concave up and the intervals where it is concave down. Show a sign graph.

Concavity of a Function

The second derivative of a function can tell us where that function is concave up and concave down. When the second derivative is positive, a function is concave up. When its's negative, the function is concave down.

Answer and Explanation:

In order to analyze the concavity of this function, we need the second derivative. We can find this second derivative after we find the fist derivative, which requires the Product Rule.

{eq}\begin{align*} f'(x) &= \frac{d}{dx} x \cdot e^{2x} + x \frac{d}{dx} e^{2x}\\ &= e^{2x} + 2xe^{2x}\\ &= e^{2x}(1+2x) \end{align*} {/eq}

We can find the second derivative by applying the Product Rule again.

{eq}\begin{align*} f''(x) &= \frac{d}{dx} e^{2x} \cdot (1+2x) + e^{2x} \frac{d}{dx} (1+2x)\\ &= 2e^{2x}(1+2x)+e^{2x}(2)\\ &= 2e^{2x}(1+2x+1)\\ &= 4e^{2x}(1+x) \end{align*} {/eq}

Let's analyze this second derivative to determine where this function is concave up and concave down. To do so, we need to determine intervals where this second derivative is either entirely positive or entirely negative, so let's set this equal to zero. As an exponential function will never equal zero, due to its horizontal asymptote, this calculation is easier than it appears.

{eq}4e^{2x}(1+x) = 0\\ 1+x = 0\\ x = -1 {/eq}

The function is therefore either entirely concave up or concave down both before and after -1. Let's test a point on either side to see which situation we're in.

{eq}f''(-2) = 4e^{2(-2)}(1+(-2)) \approx -0.0732626\\ f''(0) = 4e^{2(0)}(1+(0)) = 4 {/eq}

Therefore, this function is concave down before -1 and concave up after. We can write this in interval notation as follows.

Concave up: {eq}(-1, \infty) {/eq}

Concave down: {eq}(-\infty, -1) {/eq}

Learn more about this topic:

Concavity and Inflection Points on Graphs

from Math 104: Calculus

Chapter 9 / Lesson 5

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