French Curve

The French method of drawing curves is very systematic:

“Pratique de l’etude d’une fonction”

Let f be the function represented by the curve C


1. Simplify f(x). Determine the Domain of definition (D) of f;
2. Determine the sub-domain E of D, taking into account of the periodicity (eg. cos, sin, etc) and symmetry of f;
3. Study the Continuity of f;
4. Study the derivative of f and determine f'(x);
5. Find the limits of f within the boundary of the intervals in E;
6. Construct the Table of Variation;
7. Study the infinite branches;
8. Study the remarkable points: point of inflection, intersection points with the X and Y axes;
9. Draw the representative curve C.


\displaystyle\text{f: } x \mapsto \frac{2x^{3}+27}{2x^2}
Step 1: Determine the Domain of Definition D
D = R* = R – {0}

Step 2: There is no Periodicity and Symmetry of f
E = D = R*

[See Note below for Periodic and Symmetric example]

Step 3: Continuity of f
The function f is the quotient of 2 polynomial functions, therefore f is differentiable
=> f is continuous in ]-\infty,0[ \cup ]0,+\infty[
[See previous post CID Relation]

Step 4: Determine f’
\displaystyle\forall x \in R^{\star}, f'(x) = \frac{6x^{2}.2x^{2} - 4x (2x^{3}+27)}{4x^{4}} = \frac{4x^{4}-4.27x}{4x^{4}} = \frac{4x(x^{3}-27)}{4x^{4}}
\forall x \in R^{\star}, (x^{3} - 27 >0) \iff (x>3)
Therefore f’ has the same sign as x \mapsto x(x-3)

\begin{cases} \forall x \in ]-\infty,0[ \cup ]3,+\infty[, & f'(x)>0 \\  \forall x \in ]0,3[ , & f'(x)<0  \end{cases}

Step 5a: Limit at x=0

\displaystyle\lim_{x\to 0}(2x^{3}+27) = 27
\displaystyle\lim_{x\to 0} 2x^{2} = 0 , (\forall x \in R^{\star}, x^{2} >0)
Therefore, \displaystyle\lim_{x\to 0}f(x) = + \infty

Step 5b: Limit at x= + \infty
\displaystyle\lim_{x\to +\infty} f(x) =\lim_{x\to +\infty} \frac{2x^{3}+27}{2x^{2}}=\lim_{x\to +\infty} \frac{2x^{3}}{2x^{2}} = \lim_{x\to +\infty} x = +\infty

Step 5c: Limit at x= - \infty
\displaystyle\lim_{x\to -\infty} f(x) = \lim_{x\to -\infty} x = -\infty

Step 6: Construct the Table of Variation

\begin{array}{|l|l|l|}  \hline  x & - \infty \rightarrow \: \: \: \: 0 & 0 \:\:\:\:\: \rightarrow \:\:3 \rightarrow \:\:\: +\infty \\  \hline  f'(x) & \:\: \: \: \:\: \: + & \:\:\:\: - \:\:\:\:\:\:\:\:\: 0 \:\:\:\:\:\:\: + \\  \hline  f(x) & -\infty \nearrow +\infty & +\infty \searrow \: \frac{9}{2} \nearrow +\infty\\  \hline  \end{array}

Step 7: Study the infinite branches

7a) \displaystyle\lim_{x\to 0}f(x) = + \infty
=> y-axis is the asymptote

7b) \displaystyle\forall x \in R^{\star}, f(x) = \frac{2x^{3}+27}{2x^{2}}= x+\frac{27}{2x^{2}}
\displaystyle\lim_{x\to +\infty}\frac{27}{2x^{2}} = 0 , \displaystyle\lim_{x\to -\infty}\frac{27}{2x^{2}} = 0
\displaystyle\lim_{x\to +\infty}f(x) = x , \displaystyle\lim_{x\to -\infty}f(x) = x
=> y= x is another asymptote
\forall x \in R^{\star}, \frac{27}{2x^{2}} >0
=> The curve C is above the asymptote y=x

Step 8: Study the remarkable points: intersection points with x-axis
\forall x \in R^{\star},(2x^{3}+27 =0)  \iff (x^{3}=-\frac{27}{2})  \iff (x=-\frac{3}{\sqrt[3]{2}}) = -2.38

Step 9: Draw the representative curve C of f.

french curve

french curve

\displaystyle\text{Let g: } x \mapsto \frac{sin x}{2- cos^{2}x}
D = R
g(x) is periodic of 2π => E = [0 , 2π]
\displaystyle\forall x \in R, g(-x)= \frac{sin (-x)}{2-cos^{2}(-x)}=-\frac{sin x}{2-cos^{2}x}=-g(x)
=> g(x) is symmetric with respect to the origin point O

We can restrict our study of g(x) in E = [0,π]

\displaystyle\forall x \in R, g(\pi-x)= \frac{sin (\pi-x)}{2-cos^{2}(\pi-x)}=\frac{sin x}{2-cos^{2}x}=g(x)
=> g(x) is symmetric w.r.t. to the equation x= π/2

Finally, we can further restrict our study of g(x) in E = [0, π/2]


3 thoughts on “French Curve

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