Mathematics

Analysis of functions

Complete study of functions: domain, range (image), continuity, periodicity, and calculation of even and odd symmetries.

Domain: is the set of xx values for which the function is defined.
  • Polynomial functions: the domain is all R\mathbb{R}:
    f(x)=x2−2x+2  ⟹  Dom=R\displaystyle f(x) = x^2 - 2x + 2 \implies Dom = \mathbb{R}
  • Rational functions: the denominator cannot be 0:
    f(x)=1x2−4  ⟹  Dom=R−{−2,2}\displaystyle f(x) = \frac{1}{x^2 - 4} \implies Dom = \mathbb{R} - \{-2, 2\}
  • Irrational functions: if the root has an even index, the radicand must be greater than or equal to 0. If the root has an odd index, the domain is all R\mathbb{R}:
    f(x)=x−4  ⟹  x−4≥0  ⟹  Dom=[4,∞)\displaystyle f(x) = \sqrt{x - 4} \implies x - 4 \ge 0 \implies Dom = [4, \infty)
  • Logarithmic functions: the argument must be strictly greater than 0:
    f(x)=ln⁡(x−4)  ⟹  x−4>0  ⟹  Dom=(4,∞)\displaystyle f(x) = \ln(x - 4) \implies x - 4 > 0 \implies Dom = (4, \infty)
  • Exponential functions: the domain is all R\mathbb{R}:
    f(x)=ex−2  ⟹  Dom=R\displaystyle f(x) = e^{x-2} \implies Dom = \mathbb{R}
  • Sine and cosine functions: the domain is all R\mathbb{R}:
    f(x)=sin⁡x  ⟹  Dom=R\displaystyle f(x) = \sin x \implies Dom = \mathbb{R}
  • Tangent function: it is not defined at odd multiples of π/2\pi/2:
    f(x)=tan⁡x  ⟹  Dom=R−{π2+kπ}(k∈Z)\displaystyle f(x) = \tan x \implies Dom = \mathbb{R} - \left\{ \frac{\pi}{2} + k\pi \right\} \quad (k \in \mathbb{Z})
Range or image: is the set of values that the function takes over its entire domain.
  • If we have the analytical expression of the function, and it has an inverse, the range is the domain of the inverse function.
  • If we have the graph of the function, or can sketch it, we determine the range by projecting onto the yy-axis.
A function is said to be continuous at a point x0x_0 if the following conditions are met:
  • The image of the point x0x_0 exists (∃f(x0)\exists f(x_0))
  • The limit as x→x0x \to x_0 exists and is finite:
    lim⁡x→x0−f(x)=lim⁡x→x0+f(x)=k\displaystyle \lim_{x \to x_0^-} f(x) = \lim_{x \to x_0^+} f(x) = k
    The limit exists if both one-sided limits match.
  • Both conditions match, meaning:
    f(x0)=lim⁡x→x0−f(x)=lim⁡x→x0+f(x)\displaystyle f(x_0) = \lim_{x \to x_0^-} f(x) = \lim_{x \to x_0^+} f(x)
We say a function is periodic when its shape repeats over a certain interval called the period. E.g.:
  • Sine and cosine function: the period is 2π2\pi
  • Tangent function: the period is π\pi
Even symmetry
Symmetry with respect to the Y-axis: we check if it satisfies:
f(x)=f(−x)f(x) = f(-x)
Example: f(x)=x4−x2\displaystyle f(x) = x^4 - x^2
f(x)=x4−x2\displaystyle f(x) = x^4 - x^2
f(−x)=(−x)4−(−x)2=x4−x2\displaystyle f(-x) = (-x)^4 - (-x)^2 = x^4 - x^2

It has even symmetry.
Graph of a function with even symmetry (respect to Y-axis)
Odd symmetry
Symmetry with respect to the origin O: we check if it satisfies:
−f(x)=f(−x)-f(x) = f(-x)
Example: f(x)=x3−x\displaystyle f(x) = x^3 - x
−f(x)=−(x3−x)=−x3+x\displaystyle -f(x) = -(x^3 - x) = -x^3 + x
f(−x)=(−x)3−(−x)=−x3+x\displaystyle f(-x) = (-x)^3 - (-x) = -x^3 + x

It has odd symmetry.
Graph of a function with odd symmetry (respect to the origin)
Intercepts with the axes
  • Y-intercept: is the point where x=0x=0.
    Ex.: f(x)=x2−3x+2  ⟹  f(0)=2  ⟹  (0,2)\displaystyle f(x) = x^2 - 3x + 2 \implies f(0) = 2 \implies (0, 2)
  • X-intercepts: are the points where y=0y=0.
    Ex.: f(x)=x2−3x+2  ⟹  x2−3x+2=0  ⟹  (1,0),(2,0)\displaystyle f(x) = x^2 - 3x + 2 \implies x^2 - 3x + 2 = 0 \implies (1, 0), (2, 0)
Sign of the function
We study the sign of the function in the intervals determined by the x-intercepts and the points of discontinuity (points not in the domain).

Ex.: f(x)=x−1x2−4  ⟹  Dom=R−{−2,2}\displaystyle f(x) = \frac{x-1}{x^2-4} \implies Dom = \mathbb{R} - \{-2, 2\}
It intercepts the x-axis at x=1x=1.
Interval(−∞,−2)(-\infty, -2)(−2,1)(-2, 1)(1,2)(1, 2)(2,∞)(2, \infty)
Sign of f(x)f(x)−-++−-++
Vertical Asymptotes
We study the one-sided limits at the points of discontinuity or at the boundaries of the domain:
lim⁡x→a−f(x)=±∞andlim⁡x→a+f(x)=±∞\displaystyle \lim_{x \to a^-} f(x) = \pm\infty \quad \text{and} \quad \lim_{x \to a^+} f(x) = \pm\infty
The vertical asymptote has the equation x=ax=a.
Example: f(x)=2xx−2  ⟹  Dom=R−{2}\displaystyle f(x) = \frac{2x}{x-2} \implies Dom = \mathbb{R} - \{2\}
lim⁡x→2−2xx−2=−∞lim⁡x→2+2xx−2=+∞\displaystyle \lim_{x \to 2^-} \frac{2x}{x-2} = -\infty \quad \lim_{x \to 2^+} \frac{2x}{x-2} = +\infty

Asymptote: x=2x=2
Graph with vertical asymptote at x=2
Horizontal Asymptotes
We have horizontal asymptotes when the infinite limits of the function exist:
lim⁡x→±∞f(x)=k\displaystyle \lim_{x \to \pm\infty} f(x) = k
In rational functions if degree num ≤\le degree denom.
The horizontal asymptote has the equation y=ky=k.
Example: f(x)=2xx−2\displaystyle f(x) = \frac{2x}{x-2}
lim⁡x→−∞2xx−2=2lim⁡x→+∞2xx−2=2\displaystyle \lim_{x \to -\infty} \frac{2x}{x-2} = 2 \quad \lim_{x \to +\infty} \frac{2x}{x-2} = 2

Asymptote: y=2y=2
Graph with horizontal asymptote at y=2
Oblique (Slant) Asymptotes
There will be oblique asymptotes when the following limit exists:
lim⁡x→±∞f(x)x=m\displaystyle \lim_{x \to \pm\infty} \frac{f(x)}{x} = m
In rational functions if degree num = degree denom + 1.
It has the equation of a line y=mx+ny = mx + n, where:
m=lim⁡x→±∞f(x)xn=lim⁡x→±∞(f(x)−mx)\displaystyle m = \lim_{x \to \pm\infty} \frac{f(x)}{x} \quad n = \lim_{x \to \pm\infty} (f(x) - mx)
Example: f(x)=x2−xx−3\displaystyle f(x) = \frac{x^2 - x}{x - 3}
m=lim⁡x→±∞f(x)x=lim⁡x→±∞x2−xx2−3x=1\displaystyle m = \lim_{x \to \pm\infty} \frac{f(x)}{x} = \lim_{x \to \pm\infty} \frac{x^2 - x}{x^2 - 3x} = 1
n=lim⁡x→±∞(f(x)−x)=lim⁡x→±∞x2−x−x2+3xx−3=2\displaystyle n = \lim_{x \to \pm\infty} (f(x) - x) = \lim_{x \to \pm\infty} \frac{x^2 - x - x^2 + 3x}{x - 3} = 2

Asymptote: y=x+2y = x + 2
Graph with oblique asymptote at y=x+2
Horizontal and oblique asymptotes are mutually exclusive; that is, if one type exists, the other will not.
Parabolic Branches
There will be parabolic branches when:
lim⁡x→±∞f(x)x=∞\displaystyle \lim_{x \to \pm\infty} \frac{f(x)}{x} = \infty
Example: f(x)=x3−1\displaystyle f(x) = x^3 - 1
lim⁡x→±∞f(x)x=lim⁡x→±∞x3−1x=±∞\displaystyle \lim_{x \to \pm\infty} \frac{f(x)}{x} = \lim_{x \to \pm\infty} \frac{x^3 - 1}{x} = \pm\infty
Graph with a parabolic branch
Local maxima and minima
We look for the points where the derivative is zero:
f′(x)=0\displaystyle f'(x) = 0
The solutions, x0x_0, are candidate points for maxima and minima. They are verified using the second derivative test:
If f′′(x0)>0 it is a local minimumIf f′′(x0)<0 it is a local maximum\begin{aligned} \textit{If } f''(x_0) > 0 & \textit{ it is a local minimum} \\ \textit{If } f''(x_0) < 0 & \textit{ it is a local maximum} \end{aligned}
Monotonicity (Intervals of increase/decrease)
We study the sign of the 1st derivative in the intervals determined by the local extrema and points of discontinuity.
If f′(x0)>0f'(x_0) > 0, it is increasing. If f′(x0)<0f'(x_0) < 0, it is decreasing.

Example: f(x)=x2x2−1  ⟹  Dom=R−{−1,1}\displaystyle f(x) = \frac{x^2}{x^2 - 1} \implies Dom = \mathbb{R} - \{-1, 1\}
f′(x)=−2x(x2−1)2=0  ⟹  x=0  ⟹  critical point at x=0\displaystyle f'(x) = \frac{-2x}{(x^2 - 1)^2} = 0 \implies x = 0 \implies \text{critical point at } x = 0
Interval(−∞,−1)(-\infty, -1)(−1,0)(-1, 0)(0,1)(0, 1)(1,∞)(1, \infty)
Sign of f′(x)f'(x)++++−-−-
Monotonicityincreasesincreasesdecreasesdecreases
Inflection points
We look for the points where the 2nd derivative is zero:
f′′(x)=0\displaystyle f''(x) = 0
The solutions are candidate inflection points. They are checked in the 3rd derivative:
If f′′′(x0)≠0f'''(x_0) \neq 0, it is an inflection point.
Concavity
We study the sign of the 2nd derivative in the intervals determined by the inflection points and points of discontinuity.
If f′′(x0)>0f''(x_0) > 0, it is concave upward ∪\cup.
If f′′(x0)<0f''(x_0) < 0, it is concave downward (convex) ∩\cap.

Example: f(x)=x3−x−1  ⟹  Dom=R\displaystyle f(x) = x^3 - x - 1 \implies Dom = \mathbb{R}
f′′(x)=6x=0  ⟹  x=0\displaystyle f''(x) = 6x = 0 \implies x = 0
f′′′(0)=6≠0  ⟹  it is an inflection point\displaystyle f'''(0) = 6 \neq 0 \implies \text{it is an inflection point}
Interval(−∞,0)(-\infty, 0)(0,∞)(0, \infty)
Sign of f′′(x)f''(x)−-++
ConcavityConvex ∩\capConcave ∪\cup
If the 2nd derivative test is inconclusive, and x=ax=a is a point that satisfies:
f′(a)=0,  f′′(a)=0,…,f(n−1)(a)=0,  f(n)(a)≠0\displaystyle f'(a) = 0, \; f''(a) = 0, \dots, f^{(n-1)}(a) = 0, \; f^{(n)}(a) \neq 0
  • If nn is even and f(n)(a)<0f^{(n)}(a) < 0, there is a local maximum at x=ax=a.
  • If nn is even and f(n)(a)>0f^{(n)}(a) > 0, there is a local minimum at x=ax=a.
  • If nn is odd, there is an inflection point at x=ax=a.