Copyright

1. Find all critical points of f(x)= e^{x} cos(x) for x = [0, \pi/3]. a. x = \{ \pi/4, \pi/2 \} ...

Question:

1. Find all critical points of {eq}f(x)= e^{x} \cos(x) {/eq} for {eq}x = [ 0, \frac{\pi}{3} ] {/eq}.

a. {eq}x = \{ \frac{\pi}{4}, \frac{\pi}{2} \} {/eq}

b. {eq}x = \{ 0, \frac{\pi}{4} \} {/eq}

c. {eq}x = \frac{\pi}{4} {/eq}

d. {eq}x = 0 {/eq}

2. Given {eq}f(x) = x^{n}e^{x} {/eq}. As {eq}n {/eq} increases, which statement is true?

a. The number of possible points of inflection decreases.

b. The number of solutions to {eq}f'(x) = 0 {/eq} decreases.

c. The number of solutions to {eq}f'(x) = 0 {/eq} remains constant.

d. The number of solutions to {eq}f'(x) = 0 {/eq} increases

Critical point

In calculus 2D, a function with one independent variable has a critical point if its first derivative has roots. Remember the first derivative is the slope of the tangent line.

Answer and Explanation:


1. Find all critical points of {eq}\displaystyle f(x)= e^{x} \cos(x) \; \;\; x \; \in \; [ 0, \frac{\pi}{3} ] \\ {/eq}


The first derivative is:

{eq}\displaystyle \ f'(x)= -\sin \left( x \right) {{\rm e}^{x}}+\cos \left( x \right) {{\rm e}^{x }} \\ \displaystyle \ f'(x)= {{\rm e}^{x}} \left( \cos \left( x \right) -\sin \left( x \right) \right) \\ {/eq}

Critical point at:

{eq}\displaystyle 0= {{\rm e}^{x}} \left( \cos \left( x \right) -\sin \left( x \right) \right) \; \Leftrightarrow \; x= \frac{\pi}{4} {/eq}

The answer is:

c. {eq}x = \frac{\pi}{4} {/eq}

2. {eq}\displaystyle f(x) = x^{n}e^{x} {/eq}.

The first derivative is:

{eq}\displaystyle \ f'(x)= {\frac {{x}^{n}n{{\rm e}^{x}}}{x}}+{x}^{n}{{\rm e}^{x}} \\ \displaystyle \ f'(x)= {{\rm e}^{x}} \left( {x}^{n-1}n+{x}^{n} \right) \\ {/eq}

Critical points at

{eq}\displaystyle 0= \left( {x}^{n-1}n+{x}^{n} \right) \\ {/eq}


If {eq}n=2 {/eq}


Critical points at

{eq}\displaystyle 0= \left( {x}^{(2)-1}(2)+{x}^{(2)} \right) \\ \displaystyle 0= \left( 2x+x^2 \right) \; \Rightarrow \; 0=x(2+x) \; \Leftrightarrow \; x=0 \; \text{&} \; x=-2 \; \; \; \; \text{ two critical points } \; \\ {/eq}


If {eq}n=3 {/eq}


Critical points at

{eq}\displaystyle 0= \left( {x}^{(3)-1}(3)+{x}^{(3)} \right) \\ \displaystyle 0= \left( 3x^2+x^3 \right) \; \Rightarrow \; 0=x^2(3+x) \; \Leftrightarrow \; x=0 \; \text{&} \; x=-3 \; \; \; \; \text{ two critical points } \; \\ {/eq}


Therefore, The correct answer is:

c. The number of solutions to {eq}f'(x) = 0 {/eq} remains constant.


Learn more about this topic:

Loading...
Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
195K

Related to this Question

Explore our homework questions and answers library