7.5.2 Chapter 7 Differential Calculus

Section 7.5 Applications

7.5.1 Curve Analysis

Curve analysis is an established part of the German mathematics syllabus that consists of curve sketching for a function together with the gathering of certain standardized qualitative and quantitative information about the function graph.
Let a differentiable function $f:]a;b[\to ℝ$ with the mapping rule $x\to y=f(x)$ for $x\in ]a;b[$ be given. In this course, a complete curve analysis of $f$ includes the following information:

• maximum domain
  
• $x$- and $y$-intercepts of the graph
  
• symmetry of the graph
  
• limiting behaviour/asymptotes
  
• first derivatives
  
• extremal values
  
• monotony behaviour
  
• inflexion points
  
• bending behaviour (curvature)
  
• sketch of the graph
  

Many of these points were already discussed in Module 6. Therefore, in this section we shall only briefly repeat what is meant by the different steps of curve analysis. Subsequently, we will discuss one example of a curve analysis in detail.
The first part of the curve analysis involves the algebraic and geometric aspects of $f$:

Maximum Domain

All real numbers $x$ for which $f(x)$ exists are determined. The set $D$ of these numbers is called the maximum domain.

$x$- and $y$-Intercepts

 

• $x$-axis: All zeros of $f$ are determined.
  
• $y$-axis: The function value $f(0)$ (if $0\in D$) is calculated.
  

Symmetry of the Graph

The graph of the function is symmetrical with respect to the $y$-axis if $f(-x)=f(x)$ for all $x\in D$. Then the function $f$ is called even. If $f(-x)=-f(x)$ for all $x\in D$, the graph is centrally symmetric with respect to the origin $(0;0)$ of the coordinate system. In this case, the function is called odd.

Asymptotic Behaviour at the Domain Boundary

The limits of the function $f$ at the boundaries of its domain are investigated.

In the second part, the function is investigated analytically by means of conclusions from the first derivatives. Of course, the first and the second derivative have to be calculated first, provided they exist.

Derivatives

Calculation of the first and the second derivative (if they exist).

Extremal Values and Monotony

The necessary condition for $x$ to be an extremum (if $x\in D$ is not a boundary point of $D$), is $f\text{'}(x)=0$.
Thus, we calculate the points $x_0$ at which the derivative $f\text{'}$ takes the value zero. If at these points also the second derivative exists, then we have:

• $f\text{'}\text{'}(x_0)>0$: $x_0$ is a minimum point of $f$.
  
• $f\text{'}\text{'}(x_0)<0$: $x_0$ is a maximum point of $f$.
  

The function $f$ is monotonically increasing on that intervals of the domain on which we have $f\text{'}(x)\ge 0$. It is monotonically decreasing on that intervals where $f\text{'}(x)\le 0$.

Inflexion Points and Curvature Properties

The necessary condition for $x$ to be an inflexion point (if the second derivative $f\text{'}\text{'}$ exists), is $f\text{'}\text{'}(x)=0$
If $f\text{'}\text{'}(w_0)=0$ and $f^{(3)}(w_0)\ne 0$, then $w_0$ is an inflexion point, i.e. the bending behaviour of $f$ changes at this point.
The function $f$ is convex (left curved) on that intervals of the domain in which we have $f^{(2)}(x)\ge 0$. It is concave (right curved) on that intervals on which we have $f^{(2)}(x)\le 0$.

Sketch of the Graph

A sketch of the graph is drawn based on the information gained during the previous steps.

Detailed Example

We investigate a function $f$ defined by the mapping rule

${\displaystyle f(x)\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}\frac{4x}{x^2+2}\hspace{0.5em}.}$


Maximum Domain
The maximum domain of this function is $D_f=ℝ$ since the denominator $x^2+2\ge 2$, i.e. it is always non-zero, and hence no points have to be excluded.  
$x$- and $y$-Intercepts
The zeros of the function are the zeros of the numerator. Hence, the graph of $f$ intersects the $x$-axis only at the origin $(0;0)$ since the numerator is only zero for $x=0$. This is also the only intersection point with the $y$-axis since there we have $f(0)=0$.  
Symmetry
To investigate the symmetry we replace the argument $x$ by $(-x)$. We have

${\displaystyle f(-x)\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}\frac{4·(-x)}{(-x{)}^2+2}\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}-\frac{4x}{x^2+2}\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}-f(x)}$

for all $x\in ℝ$. Hence, the graph of the function $f$ is centrally symmetric with respect to the origin.  
Limiting Behaviour
The function is defined on the entire set of real numbers $ℝ$, so only the limiting behaviour for $x\to \infty$ and $x\to -\infty$ has to be investigated. Since $f(x)$ is a fraction of two polynomials and the denominator has a greater power than the numerator, the $x$-axis is a horizontal asymptote in both directions:

${\displaystyle \underset{x\to \pm \infty }{lim}f(x)=0\hspace{0.5em}.}$
 
Derivatives
The first two derivatives of the function are calculated from the quotient rule. For the first derivative, we have:

${\displaystyle f\text{'}(x)\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}4·\frac{1·(x^2+2)-x·2x}{(x^2+2{)}^2}\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}4·\frac{-x^2+2}{(x^2+2{)}^2}\hspace{0.5em}.}$

Taking the derivative again and simplifying the terms results in:

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{f\text{'}\text{'}(x)}& =\hfill & 4·\frac{-2x(x^2+2{)}^2-(-x^2+2)·2(x^2+2)·2x}{(x^2+2{)}^4}\hfill \\ \multicolumn{1}{c}{}& =\hfill & 4·\frac{-2x(x^2+2)-(-x^2+2)·4x}{(x^2+2{)}^3}\hfill \\ \multicolumn{1}{c}{}& =\hfill & 4·\frac{-2x^3-4x+4x^3-8x}{(x^2+2{)}^3}\hfill \\ \multicolumn{1}{c}{}& =\hfill & 4·\frac{2x^3-12x}{(x^2+2{)}^3}\hfill \\ \multicolumn{1}{c}{}& =\hfill & 8·\frac{x(x^2-6)}{(x^2+2{)}^3}\hspace{0.5em}.\hfill \end{array}}$
 
Extremal Values
The necessary condition for an extremum at $x$ is $f\text{'}(x)=0$, in this case $-x^2+2=0$. Thus, we obtain $x_1=\sqrt{2}$ and $x_2=-\sqrt{2}$. In addition, we have to investigate the behaviour of the second derivative at these points:

${\displaystyle f\text{'}\text{'}(x_1)\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}8\frac{\sqrt{2}·(2-6)}{(2+2{)}^3}<0\hspace{0.5em},\hspace{0.5em}\hspace{0.5em}f\text{'}\text{'}(x_2)\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}8\frac{(-\sqrt{2})·(2-6)}{(2+2{)}^3}>0\hspace{0.5em}.}$

Hence, $x_1$ is a maximum point and $x_2$ is a minimum point of $f$. Inserting these values for $x$ into $f$ results in the maximum point $(\sqrt{2};\sqrt{2})$ and the minimum point $(-\sqrt{2};-\sqrt{2})$ of $f$.  
Monotony Behaviour
Since $f$ is defined on the entire set of real numbers $ℝ$, the monotony behaviour can be derived from the position of the extremal points and their types: $f$ is monotonically decreasing on $]-\infty ;-\sqrt{2}[$, monotonically increasing on $]-\sqrt{2};\sqrt{2}[$, and monotonically decreasing on $]\sqrt{2};\infty [$. Monotony intervals are always given as open intervals.  
Inflexion Points
From the necessary condition $f\text{'}\text{'}(x)=0$ for $x$ to be an inflexion point, we have the equation $8x(x^2-6)=0$. Thus, $w_0=0$, $w_1=\sqrt{6}$, and $w_2=-\sqrt{6}$ are the only solutions. The polynomial in the denominator of $f\text{'}\text{'}$ is always greater than zero. Since the polynomial in the numerator has only single roots, the second derivative $f\text{'}\text{'}(x)$ changes its sign at all these points. Hence, these points are inflexion points of $f$. The coordinates of the inflexion points $(0;0)$, $(\sqrt{6};\frac{1}{2}\sqrt{6})$, $(-\sqrt{6};-\frac{1}{2}\sqrt{6})$ are determined by inserting the corresponding values for $x$ into $f$.  
Bending Behaviour
The twice-differentiable function $f$ is convex if the second derivative is greater or equal to zero. It is concave if the second derivative is less or equal to zero. Since the polynomial in the denominator of $f\text{'}\text{'}(x)$ is always positive, it is sufficient to examine the sign of the polynomial $p(x)=8x(x-\sqrt{6})(x+\sqrt{6})$ in the numerator. For $0<x<\sqrt{6}$, it is negative ($f$ is concave there). For $x>\sqrt{6}$ it is positive ($f$ is convex there). Since $f$ is centrally symmetric, it follows that $f$ is convex on the intervals $]-\sqrt{6};0[$ and $]\sqrt{6};\infty [$ and concave on the intervals $]-\infty ;-\sqrt{6}[$ and $]0;\sqrt{6}[$.  
Sketch of the Graph