Sketching the Graph of a function

Increasing/Decreasing Functions

A function is increasing when the y-value increases as the x-value increases. More formally, given any x₁, x₂ from an interval I with x₁ < x₂ (∀x₁, x₂ ∈ I, x₁ < x₂), f(x₁) < f(x₂). See 1.a. and 1.b.

A function is decreasing when the y-value decreases as the x-value increases. More formally, given any x₁, x₂ from an interval I with x₁ < x₂ (∀x₁, x₂ ∈ I, x₁ > x₂), f(x₁) < f(x₂). 1.b.  

If f’(x) > 0 for every x on some interval I (∀x ∈ I), then f(x) is increasing on this interval. If f’(x) < 0 for every x on some interval I (∀x ∈ I), then f(x) is decreasing on this interval. If f’(x) = 0 for every x on some interval I (∀x ∈ I), then f(x) is constant on this interval. In other words, the intervals where a function is increasing (or decreasing) correspond to the intervals where its derivative is positive (or negative). 1.c.

Definition. A critical point is a point in the domain of the function where the function is either not differentiable or the derivative is equal to zero. The value of the function at a critical point is called a critical value.

Concavity and inflection points

The concavity of a function is the direction in which the function opens. It is defined as whether the function opens up or down. If it opens upwards, it is called concave up. Otherwise, it is called concave down. Notice that if the function is concave up, its derivate f’ is increasing (f’’> 0). Likewise, if the function is concave down, its derivate f’ is decreasing (f’’< 0). 1.c.  

An inflection point is a point at which the curvature changes sign, i.e., where the function changes from being concave (concave downward) to convex (concave upward), or viceversa.The graph of a differentiable function has an inflection point at (x, f(x)) if and only if its first derivative f’ has an isolated extremum (a -local- minimum or maximum) at x. If a function second derivative f″(x) exists at x₀ and x₀ is an inflection point for f, then f″(x₀) = 0.

• Example: f(x) = 3x -x³. 1.e.

f’(x) = 3 -3x² = 3(1-x)(1+x). Critical points x = ±1, f(1)=2, f(-1)=-2.

• x < -1, f’(x) < 0 ⇒ f is decreasing
• -1 < x < 1, f’(x) > 0 ⇒ f is increasing
• x > 1, f’(x) < 0 ⇒ f is decreasing

Asymptotes and End Behavior of Functions

A polynomial is a function of the form f(x) = a_nxⁿ + a_n-1xⁿ⁻¹ + … + a₂x² + a₁x + a₀. The degree of a polynomial is the highest power of x in its expression. Polynomial functions are defined and continuous on all real numbers. The degree and the leading coefficient of a polynomial determine the end behavior of its graph. 1.d.

$\lim_{x \to \infty} 3x-x^{3} = -\infty$

$\lim_{x \to -\infty} 3x-x^{3} = \infty$

f’’(x) = -6x. 0 is an inflection point.

• x < 0, f’’(x) < 0 ⇒ f’ is decreasing ⇒ f is concave downward.
• x > 0, f’’(x) > 0 ⇒ f’ is increasing ⇒ f is concave upward.

Let’s plot critical points, f(1) = 2, f(-1) = -2, f(0) = 0.

Even and odd functions

Definition. A function f(x) is even if f(-x) = f(x) for all the values of x. A function f(x) is odd if f(-x) = -f(x). An even function has reflection symmetry about the y-axis. Even functions allow us to view the y-axis as a mirror, e.g., x², |x|, and cosx. Otherwise, an odd function has symmetry about the origin. Some examples are x³, sin(x), and ¹⁄_x.

f(-x) =-3x - (-x)³ = -3x +x³ = - (3x -x³) = -f(x), so f is an even function, it has symmetry about the origin. The plot is shown in Figure 1.e.

• Let’s plot ^{x + 1}⁄_{x + 2}

f’(x) = ¹⁄_{(x + 2)²} > 0 ∀ x ∈ ℝ, x ≠ -2 ⇒ f is increasing (-∞, -2) and (-2, ∞).

Critical points are those where the derivate is zero or the derivate is not defined, e.g., x = -2.

$\lim_{x \to -2^{+}}\frac{x+1}{x+2} = -\infty, \lim_{x \to -2^{-}}\frac{x+1}{x+2} = \infty$

An asymptote is a line such that the distance between the graph and the line approaches zero as one or both of the x or y coordinates tends to infinity. In other words, it is a line that the graph approaches (it gets closer and closer, but never actually reach) as it heads or goes to positive or negative infinity.

Vertical asymptotes are vertical lines (perpendicular to the x-axis) of the form x = a (where a is a constant) near which the function grows without bound. The line x = a is a vertical asymptote if: $\lim_{x \to a^{-}}f(x)=\pm\infty$ or $\lim_{x \to a^{+}}f(x)=\pm\infty$, e.g., x = -2 is a vertical asymptote of $\frac{x+1}{x+2}$.

$\lim_{x \to \pm\infty}\frac{x+1}{x+2} = 1$

Horizontal asymptotes are horizontal lines (parallel to the x-axis) that the graph of the function approaches as x → ±∞. y = c is a horizontal asymptote if the function f(x) becomes arbitrarily close to c as long as x is sufficiently large or formally: $\lim_{x \to \infty}f(x)=c$ or $\lim_{x \to -\infty}f(x)=c$, e.g., y = 1 is a horizontal asymptote of $\frac{x+1}{x+2}$.

f’’(x) = ^-2⁄_{(x + 2)³}, x ≠ -2

• f’’(x) > 0, x < -2 ⇒ concave upward
• f’’(x) < 0, x > -2 ⇒ concave downward

The plot is shown in Figure 1.a.  

Example: $\frac{x^{3}+4}{x^{2}}$

• Domain = ℝ - {0}
• $\lim_{x \to 0}\frac{x^{3}+4}{x^{2}} = \frac{4}{0^{+}} = \infty$, x = 0 is a vertical asymptote.
• $\lim_{x \to \infty}\frac{x^{3}+4}{x^{2}} = \infty, \lim_{x \to -\infty}\frac{x^{3}+4}{x^{2}} = -\infty$.

When a linear asymptote is not parallel to the x- or y-axis, it is called an oblique asymptote. A function f(x) is asymptotic to y = mx + n (m ≠ 0) if: m = $\lim_{x \to \pm\infty} \frac{f(x)}{x} = m, \lim_{x \to \pm\infty} (f(x) -mx) = n.$

• m = $\lim_{x \to \pm\infty}\frac{\frac{x^{3}+4}{x^{2}}}{x} = \lim_{x \to 0}\frac{x^{3}+4}{x^{3}}=1, lim_{x \to \pm\infty} \frac{x^{3}+4}{x^{2}} -x=lim_{x \to \pm\infty} \frac{4}{x^{3}} = 0,$ y = x is an oblique asymptote.

• f’(x) = $\frac{3x^{4}+(x^{3}+4)2x}{x^{4}}= \frac{3x^{3}-2x^{3}-8}{x^{3}}= \frac{x^{3}-8}{x^{3}}$, f’(x) = 0 ⇒ x = 2

• x < 0, f’(x) > 0 ⇒ f increasing

• 0 < x < 2, f’(x) < 0 ⇒ f decreasing

• x > 2, f’(x) > 0 ⇒ f increasing

The plot is shown in Figure 1.c.  

Example: $\frac{x^{2}}{1-x^{2}}$

• Domain = ℝ - {1, -1}
• f(-x)=f(x), so this function is even.
• $\lim_{x \to \pm\infty}\frac{x^{2}}{1-x^{2}}=-1$, y = -1 is a horizontal asymptote.
• $\lim_{x \to 1^{+}}\frac{x^{2}}{1-x^{2}}=\frac{1}{0^{-}} = -\infty, \lim_{x \to 1^{-}}\frac{x^{2}}{1-x^{2}}=\frac{1}{0^{+}} = \infty$
• $\lim_{x \to -1^{+}}\frac{x^{2}}{1-x^{2}}=\frac{1}{0^{+}} = \infty, \lim_{x \to -1^{-}}\frac{x^{2}}{1-x^{2}}=\frac{1}{0^{-}} = -\infty$, x = -1 and x = 1 are vertical asymptotes.
• f’(x) = $\frac{2x}{(1-x^{2})^{2}}$. Critical points: 0 (f’(x)=0), 1 and -1 (f is undefined).
  x < -1, f’(x) < 0 ⇒ f decreasing
  -1 < x < 0, f’(x) < 0 ⇒ f decreasing
  0 < x < 1, f’(x) > 0 ⇒ f increasing
  x > 1, f’(x) > 0 ⇒ f increasing
• f’’(x) = $\frac{6x^{2}+2}{(1-x^{2})^{3}}$ x < -1, f’’(x) < 0 ⇒ concave downward
  -1 < x < 1, f’’(x) > 0 ⇒ concave upward
  x > 1, f’’(x) < 0 ⇒ concave downward

The plot is shown in Figure 1.d.  

General strategy to plot functions

1. Find undefined values and its domain.
2. Locate discontinuities and the x- and y-intercepts.
3. Study the end behavior, determine whether f has any horizontal or vertical asymptotes.
4. Check for even or odd symmetry.
5. Calculate f’. Solve f’(x) = 0. Plot easy points, critical points and values, and determine the intervals where f is increasing or decreasing.
6. Determine whether f has any local extrema, i.e., minimums or maximums.
7. Calculate f’’. Determine the intervals where f is concave up or down, and its inflection points.

• Example. f(x) = ^x⁄_lnx, x > 0.

There is an undefined value x = 1, and $\lim_{x \to 1^{+}}\frac{x}{lnx}=\frac{1}{0^{+}}=\infty, \lim_{x \to 1^{-}}\frac{x}{lnx}=\frac{1}{0^{-}}=-\infty$. Therefore, x = 1 is a vertical asymptote.

Let’s study the end behavior: $\lim_{x \to 0^{+}}\frac{x}{lnx}=\frac{0}{-\infty}=0, \lim_{x \to \infty}\frac{x}{lnx}=\lim_{x \to \infty} {\frac{1}{\frac{1}{x}}}~ -L’Hopitals Rule-~= \lim_{x \to \infty}x = \infty$

f’(x) = $\frac{lnx-x(\frac{1}{x})}{(lnx)^{2}}=~ \frac{lnx -1}{(lnx)^{2}}$

f’(x) = 0, x = e. f(e) = e.

• 0 < x < 1, f’(x) < 0 ⇒ f is decreasing
• 1 < x < e, f’(x) < 0 ⇒ f is decreasing
• x > 0, f’(x) > 0 ⇒ f is increasing

f’’(x) = $\frac{\frac{1}{x}(lnx)^{2}-\frac{(lnx-1)2lnx}{x}}{(lnx)^{4}}=~ \frac{\frac{lnx-2(lnx-1)}{x}}{(lnx)^{3}}=~ \frac{2-lnx}{x(lnx)^{3}}$

• 0 < x < 1, f’’(x) < 0 ⇒ concave downward
• 1 < x < e², f’(x) > 0 ⇒ concave upward
• x > e², f’(x) < 0 ⇒ concave downward The plot is shown in Figure 1.b.