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Consider the function f(x)= \frac{\ln(x)}{x^6}. For this function, there are two important...

Question:

Consider the function {eq}f(x)= \frac{\ln(x)}{x^6} {/eq}. For this function, there are two important intervals: (A, B] and [B, {eq}\infty {/eq}) where A and B are critical numbers or numbers where the function is undefined. Find A Find B. For each of the following intervals, tell whether f(x) is increasing or decreasing.

(A, B] _____

[B, {eq}\infty {/eq}) _____

Sketching Graphs

To see where a function {eq}\displaystyle y=f(x) {/eq} is increasing or decreasing, we look at the sign of its derivative: {eq}\displaystyle \text{ if } f'(x)>0 \implies f(x) \text{ is increasing}\\ \displaystyle \text{ if } f'(x)<0 \implies f(x) \text{ is decreasing}. {/eq}

The points where the derivative is zero or undefined are the critical points.

Answer and Explanation:

To find the critical point of the function {eq}\displaystyle f(x)=\frac{\ln x}{x^6}, {/eq} we need to establish the domain, first.

The domain is given by {eq}\displaystyle x\neq 0 \text{ and } x>0 \implies \text{ the domain is }(0,\infty). {/eq}

To find the critical points, we need the derivative function,

{eq}\displaystyle \begin{align} f'(x)&=\frac{d}{dx}\left(\frac{\ln x}{x^6}\right)\\ &=\frac{\frac{1}{x}\cdot x^6-6x^5\cdot \ln x}{x^{12}}\\ &=\frac{x^5-6x^5\ln x}{x^{12}}\\ &=\frac{1-6\ln x}{x^{7}}, \text{by simplifying by } x^5 \neq 0\\ \end{align} {/eq}

So, the derivative function is undefined when the denominator is 0, but in the domain {eq}\displaystyle (0,\infty) {/eq} the denominator is never zero,

and the derivative is zero when {eq}\displaystyle \ln x=\frac{1}{6} \implies x=e^{1/6}, {/eq}

Therefore, the function is not defined at {eq}\displaystyle \boxed{A=0}, {/eq} and has a critical point at {eq}\displaystyle \boxed{B=e^{1/6}}. {/eq}

The function is increasing when the first derivative is positive, {eq}\displaystyle f'(x)=\frac{1-6\ln x}{x^{7}}>0 \text{ on } (0,\infty) \text{ when }\\ \displaystyle 1-6\ln x>0\iff \ln x<\frac{1}{6} \iff x\in\left(0,e^{1/6}\right). {/eq}

Therefore, {eq}\displaystyle \boxed{\text{ the function is increasing on } \left(0,e^{1/6}\right] \text{ and is decreasing on } \left[e^{1/6}, \infty\right)}. {/eq}


Learn more about this topic:

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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
195K

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