3. Analysis of a real function

Analysis of a real function

We will find and study:

  • the domain and zeroes
  • continuity, limits in the points in which function is not continuous and at the ends of intervals,
  • asymptotes,
  • differentiability and derivatives,
  • intervals of monotonicity and extrema
  • second derivative, convexity, inflection points
  • the table of the function,
  • parity, periodicity,
  • sketch of the graph,
  • range.

An example

We will study g(x)=\frac{x^3}{1-x^2}.

the domain and zeroes

The denominator 1-x^2=0, if x\in\{-1,1\}. Therefore, D_f=\mathbb{R}\setminus\{-1,1\}. Moreover, g(x)=0 iff x=0.

Continuity, limits in the points in which function is not continuous and at the ends of intervals,

The function is continuous on (-\infty,-1),(-1,1) and (1,\infty).

Furthermore:

    \[\lim_{x\to-\infty}g(x)=\infty\]

    \[\lim_{x\to-1^-}g(x)=\infty\]

    \[\lim_{x\to-1^+}g(x)=-\infty\]

    \[\lim_{x\to 1^-}g(x)=\infty\]

    \[\lim_{x\to 1^+}g(x)=-\infty\]

    \[\lim_{x\to\infty}g(x)=-\infty\]

Asymptotes

Therefore we have vertical asymptotes x=-1 i x=1. There are no horizontal asymptotes.

We check oblique asymptotes:

    \[\lim_{x\to \infty} \frac{g(x)}{x}=\lim_{x\to\infty} \frac{x^2}{1-x^2}=\lim_{x\to\infty} \frac{1}{1/x^2-1}=-1.\]

    \[\lim_{x\to \infty} g(x)+x=\lim_{x\to\infty}\frac{x^3+x-x^3}{1-x^2}=\lim_{x\to\infty} \frac{1}{1/x-x}=0.\]

Therefore, y=-x is the right oblique asymptote.

    \[\lim_{x\to -\infty} \frac{g(x)}{x}=\lim_{x\to-\infty} \frac{x^2}{1-x^2}=\lim_{x\to-\infty} \frac{1}{1/x^2-1}=-1.\]

    \[\lim_{x\to -\infty} g(x)+x=\lim_{x\to-\infty}\frac{x^3+x-x^3}{1-x^2}=\lim_{x\to-\infty} \frac{1}{1/x-x}=0.\]

Therefore, y=-x is also the left oblique asymptote.

Differentiability and derivatives

The funtion in differentiable on the whole domain and:

    \[g'(x)=\dfrac{3x^2(1-x^2)+2x^4}{(1-x^2)^2}=-\dfrac{x^2(x^2-3)}{(1-x^2)^2}.\]

Intervals of monotonicity and extrema

x^2(x^2-3)=0 if x=0 or x=\pm \sqrt{3}. Therefore:

  • on (-\infty,-\sqrt{3}) we have g'(x)<0, so f decreases,
  • on (-\sqrt{3},-1) we have g'(x)>0, so f increases,
  • on (-1,0) we have g'(x)>0, so f increases,
  • on (0,1) we have g'(x)>0, so f increases,
  • on (1,\sqrt{3}) we have g'(x)>0, so f increases,
  • on (\sqrt{3},\infty) we have g'(x)<0, so f decreases.

Therefore, in x=-\sqrt{3} the function has its local minimum, and in x=\sqrt{3} its local maximum.

Second derivative, convexity, inflection points

    \[g''(x)=\left(\dfrac{x^2(x^2-3)}{(1-x^2)^2}\right)'=-\dfrac{2x(x^2+3)}{(1-x^2)^3},\]

  • on (-\infty,-1) we have g''(x)>0, so f is convex,
  • on (-1,0) we have g''(x)<0, so f is concave,
  • on (0,1) we have g''(x)>0, so f is convex,
  • on (1,\infty) we have g''(x)<0, so f is concave.

Therefore 0 is an inflection point.

The table of the function

g' g'' g
(-\infty,-\sqrt{3}) <0 >0 decreasing, convex
-\sqrt{3} =0 >0 local minimum
(-\sqrt{3},-1) >0 >0 increasing, convex
-1 vertical asymptote
(-1,0) >0 <0 increasing, concave
0 =0 =0 inflection point
(0,1) >0 >0 increasing, convex
1 vertical asymptote
(1,\sqrt{3}) >0 <0 increasing, concave
\sqrt{3} =0 <0 local maximum
(\sqrt{3},\infty) <0 <0 decreasing, wklęsła.

parity, periodicity

The function is odd, since f(-x)=\frac{-x^3}{(1-x^2)}=-f(x). Therefore it is not even, because it is not a constant zero function. Obviously, it is not a periodic function.

Sketch of the graph

Range

Obviously, R_g=\mathbb{R}.