Chapter 4, Problem 4RWDT

How could you complete an algebraic justification that the function $f(x)=-\frac{4x}{x^2+1}$ has an absolute maximum at $(-1,2)$ and an absolute minimum at $(1,-2)$ ?

To determine

To calculate: An algebraic justification that the function $f(x)=-\frac{4x}{x^2+1}$ has an absolute maximum at $(-1,2)$ and an absolute minimum at $(1,-2)$

Answer to Problem 4RWDT

JUSTIFIED - $f(x)=-\frac{4x}{x^2+1}$ has an absolute maximum at $(-1,2)$ and an absolute minimum at $(1,-2)$

Explanation of Solution

Given information: An algebraic justification that the function $f(x)=-\frac{4x}{x^2+1}$ has an absolute maximum at $(-1,2)$ and an absolute minimum at $(1,-2)$

Calculation:

To justify that a function has an absolute maximum and an absolute minimum

We need to find the domain and range of the function, $f(x)=-\frac{4x}{x^2+1}$

Domain of $f(x)=-\frac{4x}{x^2+1}$ : $[Solution:-\infty <x<\infty ]$

  $[Interval$ $Notation:(-\infty ,\infty )$ $]$

(The function has no undefined points nor domain constraints. Therefore, the domain is $-\infty <x<\infty$ )

Range of $f(x)=-\frac{4x}{x^2+1}$ : $[Solution:-2\le f(x)<0$ or $0<f(x)<2]$

  $[$ $Interval$ $Notation:[-2,0)\cup (0,2]]$

The axis interception points of $f(x)=-\frac{4x}{x^2+1}$ : X Intercepts: $(0,0)$ ,Y Intercepts: $(0,0)$

Vertical asymptotes of $f(x)=-\frac{4x}{x^2+1}$ : None

Horizontal asymptotes of $f(x)=-\frac{4x}{x^2+1}$ : $y=0$

Thus, Extreme points of $f(x)=-\frac{4x}{x^2+1}$ : Maximum $(-1,2)$ , Minimum $(1,-2)$