Determine the intervals where the function f(x) = x\sqrt{9 - x^2} is concave up or down.

Question:

Determine the intervals where the function {eq}f(x) = x\sqrt{9 - x^2} {/eq} is concave up or down.

Concavity and Point of Inflection:

{eq}\text{The graph of y = f (x) is concave upward on those intervals where y = f "(x) > 0. The graph of y = f (x) is concave downward on those intervals where y = f "(x) < 0.} \\ \text{If the graph of y = f (x) has a point of inflection then} y = f ''(x) = 0. {/eq}

Answer and Explanation:

$$\text{To check the concavity, we need to evaluate f''(x) and put it equal to 0}\\ f(x)=x\sqrt{9 - x^2}\\ \text{using product rule}\\ f'(x)\;=\;1\cdot \sqrt{9-x^2}+x\cdot \frac{1}{2}(9-x^2)^{-1/2}(-2x)\\ \text{On simplifying furthur}\\ f'(x)=\sqrt{9-x^2}-\frac{x^2}{\sqrt{9-x^2}}\\ f'(x)=\frac{-2x^2+9}{\sqrt{9-x^2}}\\ \text{Thus on double differentiation we will get: }\\ \text{using product rule}\\ f''(x)\;=\; \frac{\sqrt{9-x^2}(-4x)-(-2x^2+9).(1/2)(9-x^2)^{-1/2}(-2x)}{9-x^2}\\ \text{On simplifying furthur}\\ f''(x)=\frac{-4x\sqrt{9-x^2}+\frac{x(-2x^2+9)}{\sqrt{9-x^2}}}{9-x^2}\\ f''(x)=\frac{-4x(9-x^2)+x(-2x^2+9)}{(9-x^2)^{3/2}}\\ f''(x)=\frac{x(2x^2-27)}{(-x^2+9)^{3/2}}\\ \text{Now, equate }f''(x)=0\\ \frac{x(2x^2-27)}{(-x^2+9)^{3/2}}=0\\ x(2x^2-27)=0\\ x=0,\sqrt{\frac{27}{2}}\\ \text{At x= }\pm3,\; f''(x) \text{is not defined and }\pm3\text{ won't lie in domain of function as well ,& at -3>x>0, }f''(x)>0 \text{ and at 0>x>3, f''(x)<0}\\ \text{So the function is concave up in the interval } [-3,0) \text{ and concave down in the interval} (0,3]\\ $$


Learn more about this topic:

Loading...
Understanding Concavity and Inflection Points with Differentiation

from Math 104: Calculus

Chapter 10 / Lesson 6
14K

Related to this Question

Explore our homework questions and answers library