consider the function f(x)=4x + 3x^(-1). for this function there are four important intervals…

consider the function f(x)=4x + 3x^(-1). for this function there are four important intervals: (-∞,a,a,b),(b,c, and c,∞) where a, and c are the critical numbers and the function is not defined at b. find a and b and c for each of the following open intervals, tell whether f(x) is increasing or decreasing. (-∞,a):___ (a,b):___ (b,c):___ (c,∞)___ note that this function has no inflection points, but we can still consider its concavity. for each of the following intervals, tell whether f(x) is concave up or concave down. (-∞,b):___ (b,∞):___

consider the function f(x)=4x + 3x^(-1). for this function there are four important intervals: (-∞,a,a,b),(b,c, and c,∞) where a, and c are the critical numbers and the function is not defined at b. find a and b and c for each of the following open intervals, tell whether f(x) is increasing or decreasing. (-∞,a):___ (a,b):___ (b,c):___ (c,∞)___ note that this function has no inflection points, but we can still consider its concavity. for each of the following intervals, tell whether f(x) is concave up or concave down. (-∞,b):___ (b,∞):___

Answer

Explanation:

Step1: Find the derivative of the function

The function is $f(x)=4x + 3x^{-1}$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=4-3x^{-2}=4-\frac{3}{x^{2}}=\frac{4x^{2}-3}{x^{2}}$.

Step2: Find the critical numbers

Set $f'(x) = 0$, then $\frac{4x^{2}-3}{x^{2}}=0$. Since the denominator $x^{2}\neq0$ for the domain of $f'(x)$, we solve $4x^{2}-3 = 0$. So $x^{2}=\frac{3}{4}$, and $x=\pm\frac{\sqrt{3}}{2}$. The function $f(x)=4x + 3x^{-1}$ is not defined at $x = 0$. So $A=-\frac{\sqrt{3}}{2}$, $B = 0$, $C=\frac{\sqrt{3}}{2}$.

Step3: Determine the increasing and decreasing intervals

We use test - points in the intervals $(-\infty,-\frac{\sqrt{3}}{2})$, $(-\frac{\sqrt{3}}{2},0)$, $(0,\frac{\sqrt{3}}{2})$ and $(\frac{\sqrt{3}}{2},\infty)$. For $x=-1\in(-\infty,-\frac{\sqrt{3}}{2})$, $f'(-1)=4 - 3=1>0$, so $f(x)$ is increasing on $(-\infty,-\frac{\sqrt{3}}{2})$. For $x =-\frac{1}{2}\in(-\frac{\sqrt{3}}{2},0)$, $f'(-\frac{1}{2})=4-12=-8<0$, so $f(x)$ is decreasing on $(-\frac{\sqrt{3}}{2},0)$. For $x=\frac{1}{2}\in(0,\frac{\sqrt{3}}{2})$, $f'(\frac{1}{2})=4 - 12=-8<0$, so $f(x)$ is decreasing on $(0,\frac{\sqrt{3}}{2})$. For $x = 1\in(\frac{\sqrt{3}}{2},\infty)$, $f'(1)=4 - 3=1>0$, so $f(x)$ is increasing on $(\frac{\sqrt{3}}{2},\infty)$.

Step4: Find the second - derivative and concavity

$f'(x)=4-\frac{3}{x^{2}}$, then $f''(x)=\frac{6}{x^{3}}$. For $x=-1\in(-\infty,0)$, $f''(-1)=-6<0$, so $f(x)$ is concave down on $(-\infty,0)$. For $x = 1\in(0,\infty)$, $f''(1)=6>0$, so $f(x)$ is concave up on $(0,\infty)$.

Answer:

$A=-\frac{\sqrt{3}}{2}$ $B = 0$ $C=\frac{\sqrt{3}}{2}$ $(-\infty,A)$: Increasing $(A,B)$: Decreasing $(B,C)$: Decreasing $(C,\infty)$: Increasing $(-\infty,B)$: Concave down $(B,\infty)$: Concave up