Question

1. the function f and its first and second derivatives are given:

$f(x) = \frac{3(x^2 - 1)}{x^2 + 3}$, $f(x) = \frac{24x}{(x^2 + 3)^2}$, $f(x) = \frac{-72(x^2 - 1)}{(x^2 + 3)^3}$,
obtain the following information.
(2) (a) the intercepts:__________.
(2) (b) horizontal asymptotes: _______. vertical asymptotes: none.
(3) (c) coordinates of local maxima: ______, local minima: ______.
(3) (d) coordinates of the inflection points: ________
(2) (e) use this information to sketch the graph of f
graph with x-axis from -6 to 6 and y-axis from -2 to 4, with axes labeled and grid lines (partial)

Explanation:

Part (a): Find Intercepts

Step 1: Find x - intercepts

To find the x - intercepts, we set \(y = f(x)=0\). So we solve the equation \(\frac{3(x^{2}-1)}{x^{2}+3}=0\). A fraction is zero when its numerator is zero (and the denominator is not zero). The denominator \(x^{2}+3\) is always positive for all real \(x\) (since \(x^{2}\geq0\), so \(x^{2}+3\geq3>0\)). We solve \(3(x^{2}-1)=0\), which simplifies to \(x^{2}-1 = 0\), or \((x - 1)(x + 1)=0\). So \(x=1\) or \(x=-1\). The x - intercepts are \((-1,0)\) and \((1,0)\).

Step 2: Find y - intercept

To find the y - intercept, we set \(x = 0\) in \(f(x)\). Substitute \(x = 0\) into \(f(x)=\frac{3(0^{2}-1)}{0^{2}+3}=\frac{3(-1)}{3}=-1\). So the y - intercept is \((0,-1)\).

Part (b): Find Horizontal Asymptotes

Step 1: Recall the formula for horizontal asymptotes

For a rational function \(y=\frac{ax^{n}+...}{bx^{m}+...}\), if \(n = m\), the horizontal asymptote is \(y=\frac{a}{b}\). For \(f(x)=\frac{3x^{2}-3}{x^{2}+3}\), the degree of the numerator \(n = 2\) and the degree of the denominator \(m = 2\). Here, \(a = 3\) (coefficient of \(x^{2}\) in the numerator) and \(b = 1\) (coefficient of \(x^{2}\) in the denominator). So the horizontal asymptote is \(y=\frac{3}{1}=3\).

Part (c): Find Local Maxima and Minima

Step 1: Find critical points

To find critical points, we set \(f^{\prime}(x)=0\) or find where \(f^{\prime}(x)\) is undefined. The derivative \(f^{\prime}(x)=\frac{24x}{(x^{2}+3)^{2}}\). The denominator \((x^{2}+3)^{2}\) is never zero for real \(x\) (since \(x^{2}+3>0\)), so we set the numerator equal to zero: \(24x=0\), which gives \(x = 0\).

Step 2: Use the second - derivative test

We use the second - derivative \(f^{\prime\prime}(x)=\frac{-72(x^{2}-1)}{(x^{2}+3)^{3}}\). Evaluate \(f^{\prime\prime}(x)\) at \(x = 0\): \(f^{\prime\prime}(0)=\frac{-72(0^{2}-1)}{(0^{2}+3)^{3}}=\frac{-72(-1)}{27}=\frac{72}{27}=\frac{8}{3}>0\). When \(f^{\prime\prime}(c)>0\) at a critical point \(x = c\), the function has a local minimum at \(x = c\). Now, find \(f(0)=\frac{3(0^{2}-1)}{0^{2}+3}=-1\). So the local minimum is at \((0,-1)\). Since the first - derivative \(f^{\prime}(x)\) changes sign from negative to positive at \(x = 0\) (for \(x<0\), \(f^{\prime}(x)<0\) because \(x<0\) and numerator \(24x<0\); for \(x>0\), \(f^{\prime}(x)>0\) because \(x>0\) and numerator \(24x>0\)), and there is only one critical point, there is no local maximum? Wait, wait, maybe I made a mistake. Wait, let's re - examine the first - derivative. \(f^{\prime}(x)=\frac{24x}{(x^{2}+3)^{2}}\). The denominator is always positive. So when \(x<0\), \(f^{\prime}(x)<0\) (function is decreasing), when \(x>0\), \(f^{\prime}(x)>0\) (function is increasing). So the function has a local minimum at \(x = 0\) and no local maximum? Wait, but let's check the original function. \(f(x)=\frac{3(x^{2}-1)}{x^{2}+3}=\frac{3x^{2}-3}{x^{2}+3}=3-\frac{12}{x^{2}+3}\). As \(x\to\pm\infty\), \(f(x)\to3\). The function is symmetric about the y - axis? Let's check \(f(-x)=\frac{3((-x)^{2}-1)}{(-x)^{2}+3}=\frac{3(x^{2}-1)}{x^{2}+3}=f(x)\), so it is an even function. So the graph is symmetric about the y - axis. So the function decreases on \((-\infty,0)\) and increases on \((0,\infty)\), so only a local minimum at \((0,-1)\) and no local maximum.

Part (d): Find Inflection Points

Step 1: Set \(f^{\prime\prime}(x)=0\)

We set \(f^{\prime\prime}(x)=\frac{-72(x^{2}-1)}{(x^{2}+3)^{3}}=0\). The denominator is never zero, so we solve \(-72(x^{2}-1)=0\), which gives \(x^{2}-1 = 0\), so \(x = 1\) or \(x=-1\).

Step 2: Check the sign of \(f^{\prime\prime}(x)\) around \(x = 1\) and \(x=-1\)

For \(x<-1\), let's take \(x=-2\). \(f^{\prime\prime}(-2)=\frac{-72((-2)^{2}-1)}{((-2)^{2}+3)^{3}}=\frac{-72(4 - 1)}{(4 + 3)^{3}}=\frac{-72\times3}{343}<0\). For \(-1<x<1\), let's take \(x = 0\). \(f^{\prime\prime}(0)=\frac{-72(0 - 1)}{(0 + 3)^{3}}=\frac{72}{27}>0\). For \(x>1\), let's take \(x = 2\). \(f^{\prime\prime}(2)=\frac{-72(4 - 1)}{(4 + 3)^{3}}=\frac{-72\times3}{343}<0\). So the concavity changes at \(x=-1\) and \(x = 1\). Now, find \(f(-1)\) and \(f(1)\). We know \(f(-1)=0\) and \(f(1)=0\). So the inflection points are \((-1,0)\) and \((1,0)\).

Part (e): Sketch the Graph

- **Intercepts**: \((-1,0)\), \((1,0)\), \((0,-1)\)
- **Asymptotes**: Horizontal asymptote \(y = 3\), no vertical asymptotes.
- **Local Extrema**: Local minimum at \((0,-1)\), no local maximum.
- **Inflection Points**: \((-1,0)\) and \((1,0)\)
- **Behavior**: The function is even (symmetric about the y - axis). For \(x\in(-\infty,-1)\), the function is concave down and decreasing (since \(f^{\prime}(x)<0\) for \(x<0\) and \(f^{\prime\prime}(x)<0\) for \(x<-1\)). For \(x\in(-1,0)\), the function is concave up and decreasing ( \(f^{\prime}(x)<0\) and \(f^{\prime\prime}(x)>0\)). For \(x\in(0,1)\), the function is concave up and increasing ( \(f^{\prime}(x)>0\) and \(f^{\prime\prime}(x)>0\)). For \(x\in(1,\infty)\), the function is concave down and increasing ( \(f^{\prime}(x)>0\) and \(f^{\prime\prime}(x)<0\)). As \(x\to\pm\infty\), \(f(x)\to3\).

Answers:

(a) x - intercepts: \(\boldsymbol{(-1,0),(1,0)}\); y - intercept: \(\boldsymbol{(0,-1)}\)
(b) Horizontal asymptote: \(\boldsymbol{y = 3}\)
(c) Local maxima: \(\boldsymbol{\text{None}}\); Local minima: \(\boldsymbol{(0,-1)}\)
(d) Inflection points: \(\boldsymbol{(-1,0),(1,0)}\)
(e) Sketch the graph using the above - mentioned features (intercepts, asymptotes, extrema, inflection points, and concavity/monotonicity).

Answer:

s:
(a) x - intercepts: \(\boldsymbol{(-1,0),(1,0)}\); y - intercept: \(\boldsymbol{(0,-1)}\)
(b) Horizontal asymptote: \(\boldsymbol{y = 3}\)
(c) Local maxima: \(\boldsymbol{\text{None}}\); Local minima: \(\boldsymbol{(0,-1)}\)
(d) Inflection points: \(\boldsymbol{(-1,0),(1,0)}\)
(e) Sketch the graph using the above - mentioned features (intercepts, asymptotes, extrema, inflection points, and concavity/monotonicity).