Example
The example is available in following formats:
- this HTML page
- file in Lambda:
analysis_en.lambda
- file in MS Word:
analysis_en.doc
Analyse the function
f(x) =\frac{ x^3 }{ x^2 -1 } .
Solution
1. Domain of the function
x^2 -1 \neq 0
x^2 \neq 1
x \neq\pm 1
D_f =(-\infty, -1) \cup (-1, 1) \cup (1, \infty)
-1 and 1 are the points of discontinuity.
2. Even or odd function
f(-x) =\frac{ (-x)^3}{ (-x)^2 -1}=-\frac{ x^3}{ x^2 -1} =-f(x)
The function is odd.
3. Graph of the function below or above the axis x
Solving the equation f(x) =0:
\frac{ x^3}{ x^2 -1} =0 \Leftrightarrow x =0
x \in (-\infty, -1): -
x \in (-1, 0): +
x \in (0, 1): -
x \in (1, \infty): -
4. Stationary points
Solving the equation f'(x) =0:
f'(x) =\frac{ 3x^2 *(x^2 -1) -x^3 *2x}{ (x^2 -1)^2}
=\frac{ x^4 -3x^2}{ (x^2 -1)^2}
=\frac{ x^2 *(x^2 -3)}{ (x^2 -1)^2}
x =0
x^2 -3 =0
x^2 =3
x =\pm\sqrt{ 3}
x \in (-\infty, -\sqrt{ 3}): +, \uparrow
x \in (-\sqrt{ 3}, 0): -, \downarrow
x \in (0, \sqrt{ 3}): -, \downarrow
x \in (-\infty, -\sqrt{ 3}): +, \uparrow
The local maximum: [-\sqrt{ 3}, \frac{ -3\sqrt{ 3}}{ 2}]
The local minimum: [\sqrt{ 3}, \frac{ 3\sqrt{ 3}}{ 2}]
5. Inflection points
Solving the equation f''(x) =0:
f''(x) =\frac{ (4x^3 -6x) *(x^2 -1)^2 -(x^4 -3x^2) *2(x^2 -1) *2x}{ (x^2 -1)^4}
=\frac{ 2x *(x^2 -1) *[(2x^2 -3) *(x^2 -1) -2(x^4 -3x^2)]}{ (x^2 -1)^4}
=\frac{ 2x *[(2x^2 -3) *(x^2 -1) -2(x^4 -3x^2)]}{ (x^2 -1)^3}
=\frac{ 2x *[2x^4 -3x^2 -2x^2 +3 -2x^4 +6x^2]}{ (x^2 -1)^3}
=\frac{ 2x *[x^2 +3]}{ (x^2 -1)^3}
2x =0 \Rightarrow x =0
x^2 +3\neq0
x \in (-\infty, -1): -, A
x \in (-1, 0): +, V
x \in (0, 1): -, A
x \in (1, \infty): +, V
The inflection point is [0, 0].
6. Asymptotes
Oblique asymptote y =kx +q:
k =\lim_{ x \rightarrow \pm\infty}\frac{ f(x)}{ x}
=\lim_{ x \rightarrow \pm\infty}\frac{ x^3}{ x^2 -1} :x
=\lim_{ x \rightarrow \pm\infty}\frac{ x^3}{ x *(x^2 -1)}
=\lim_{ x \rightarrow \pm\infty}\frac{ x^2}{ x^2 -1}
=1
q =\lim_{ x \rightarrow \pm\infty}(f(x) -kx)
=\lim_{ x \rightarrow \pm\infty}(\frac{ x^3}{ x^2 -1} -x)
=\lim_{ x \rightarrow \pm\infty}(\frac{ x^3 -x(x^2 -1)}{ x^2 -1})
=\lim_{ x \rightarrow \pm\infty}\frac{ x}{ x^2 -1}
=0
So the oblique asymptote is y =x.
Vertical asymptotes:
- \lim_{ x \rightarrow -1^-}\frac{ x^3}{ x^2 -1}=\lim_{ x \rightarrow -1^-}\frac{ x^3}{ x +1} *\frac{ 1}{ x -1} =-\infty
- \lim_{ x \rightarrow -1^+}\frac{ x^3}{ x^2 -1}=\lim_{ x \rightarrow -1^+}\frac{ x^3}{ x +1} *\frac{ 1}{ x -1} =\infty
- \lim_{ x \rightarrow 1^-}\frac{ x^3}{ x^2 -1}=\lim_{ x \rightarrow 1^-}\frac{ x^3}{ x +1} *\frac{ 1}{ x -1} =-\infty
- \lim_{ x \rightarrow 1^+}\frac{ x^3}{ x^2 -1}=\lim_{ x \rightarrow 1^+}\frac{ x^3}{ x +1} *\frac{ 1}{ x -1} =\infty
There are two vertical asymptotes: x =-1 and x =1.
7. Final description of the graph
The function is defined for x \in (-\infty, -1) \cup (-1, 1) \cup (1, \infty).
The graph is limited by the vertical asymptotes x =-1 and x =1
and the oblique asymptote y =x.
There are two local extrema, local maximum: [-\sqrt{ 3}, \frac{ -3\sqrt{ 3}}{ 2}]
and local minimum: [\sqrt{ 3}, \frac{ 3\sqrt{ 3}}{ 2}].
The graph of the function is for x \in (-\infty, -1)
below the axis x, there is the local maximum [-\sqrt{ 3}, \frac{ -3\sqrt{ 3}}{ 2}],
the curve concaves downward, and its position is on the left side of the asymptote x =-1.
Because the point [-\sqrt{ 3}, -\sqrt{ 3}]
lying on the asymptote y =x is above the local maximum [-\sqrt{ 3}, \frac{ -3\sqrt{ 3}}{ 2}],
the whole part of the graph is
for x \in (-\infty, -1) bellow the asymptote y =x.
The graph of the function on the interval (-1, 1)
intersects with the axis x at the point [0, 0] – the inflection point, where the function
changes its shape, for x \in (-1, 0) concaves upward,
for x \in (0, 1) concaves downward, the function is odd.
The part of the graph for x \in (-1, 1)
is located between the vertical asymptotes x =-1 and x =1,
the oblique asymptote y =x intersects the graph of the function at the point [0, 0].
The graph of the function is for x \in (1, \infty)
above the axis x, there is the local minimum [\sqrt{ 3}, \frac{ 3\sqrt{ 3}}{ 2}],
the curve concaves upward, and its position is on the right side of the asymptote x =1.
Because the point [\sqrt{ 3}, \sqrt{ 3}]
lying on the asymptote y =x is below the local minimum [\sqrt{ 3}, \frac{ 3\sqrt{ 3}}{ 2}],
the whole part of the graph is
for x \in (1, \infty) above the asymptote y =x.