Sketching Graphs of Logarithmic Functions

We finish the analysis unit by discussing sketching logarithmic functions. Recall that to sketch a function we need:

Domain
Axes Intercepts
Asymptotes (horizontal and vertical)
Extreme Points (and the first derivative)
Inflection Points (and the second derivative)

Table of Contents

Asymptotes of the Logarithmic Function

A horizontal asymptote exists if the limit $\lim_{x\to\pm\infty} f(x)$ exists and is finite. We may not be able to find one of the limits (either $x\to\infty$ or $x\to-\infty$ ) depending on the domain. We may also have examples where neither of them can be found.

The vertical asymptote is studied at points of discontinuity. In other words, if we have some $\ln(f(x))$ we study the limits as $f(x)\to\infty$ or $f(x)\to0$ . Either of these could result in a vertical asymptote. This is because:

$\lim \ln(0^+)=-\infty\text{ and }\lim\ln(\infty)=\infty$

Example: Determine the equations of all vertical and horizontal asymptotes for the composite logarithmic function $f(x) = \ln\left(\frac{x-2}{x+3}\right)$ .

Sketching Logarithmic Functions

As we did when sketching general functions, we find everything in the list above and combine it to create the graph.

Example: Conduct a complete curve sketching analysis for the logarithmic function $f(x) = \ln(3x + 9)$ and include a diagram with its graph.

As you can see, logarithmic functions of the form $\ln(ax+b)$ are not too difficult to sketch. They are monotonic and their asymptotes are easy to understand. If we want to push our luck, we could take a rational function.

Example: Conduct a complete curve sketching analysis for the function $f(x) = \ln\left(\frac{x+1}{x-2}\right)$ . Identify its domain, axes intercepts, horizontal and vertical asymptotes, local extreme points, and points of inflection and draw a sketch of the curve.

Step 1: Determine the domain and axes intercepts
The argument of a natural logarithm must be strictly positive ( $> 0$ ):

$\frac{x+1}{x-2} > 0$

The critical boundary points occur where the numerator or denominator vanishes ( $x = -1$ and $x = 2$ ). Performing an interval sign test shows the quotient is positive when both linear factors share the identical sign:

$\text{Domain: } (-\infty, -1) \cup (2, \infty)$

To locate crossings along the coordinate axes:

$y$ -intercept: Evaluating $f(0) = \ln\left(\frac{0+1}{0-2}\right) = \ln\left(-\frac{1}{2}\right)$ , which is completely undefined. The graph has no $y$ -intercept because $x=0$ lies outside the allowed domain.
$x$ -intercepts: Set $f(x) = 0$ :
$\ln\left(\frac{x+1}{x-2}\right) = 0 \implies \frac{x+1}{x-2} = e^0 \implies \frac{x+1}{x-2} = 1$

$x + 1 = x - 2 \implies 1 = -2 \quad \text{(No Real Solution)}$
The graph has no $x$ -intercepts and never touches the horizontal axis.

Step 2: Find the asymptotes

Vertical Asymptotes: These occur at the open boundary thresholds of the domain where the inner argument approaches or grows infinitely large:
- As $x \to -1^-$ : The argument approaches $\frac{0^-}{-3} = 0^+ \implies \displaystyle\lim_{x \to -1^-} \ln\left(\frac{x+1}{x-2}\right) = -\infty$ .
- As $x \to 2^+$ : The argument approaches $\frac{3}{0^+} = \infty \implies \displaystyle\lim_{x \to 2^+} \ln\left(\frac{x+1}{x-2}\right) = \infty$ .
This confirms two distinct vertical asymptotes along the lines $x = -1$ and $x = 2$ .
Horizontal Asymptote: Evaluate the infinite limit profiles as $x \to \pm\infty$ :
$\lim_{x \to \pm\infty} \left(\frac{x+1}{x-2}\right) = \lim_{x \to \pm\infty} \left(\frac{1 + \frac{1}{x}}{1 - \frac{2}{x}}\right) = 1$
Passing this result inside the continuous natural logarithm function yields:
$\lim_{x \to \pm\infty} \ln\left(\frac{x+1}{x-2}\right) = \ln(1) = 0$
This establishes a horizontal asymptote along the line $y = 0$ (the $x$ -axis) on both tails.

Step 3: Extreme Points (First Derivative Analysis)
Rewrite the function via standard log laws as $f(x) = \ln(x+1) - \ln(x-2)$ to simplify calculation:

$f'(x) = \frac{1}{x+1} - \frac{1}{x-2} = \frac{(x-2) - (x+1)}{(x+1)(x-2)} = \frac{-3}{(x+1)(x-2)}$

Setting $f'(x) = 0 \implies -3 = 0$ yields no solution. Over the allowed domain intervals, the factors $(x+1)$ and $(x-2)$ always share matching signs, forcing their product to be strictly positive. Due to the negative numerator, the derivative is strictly negative ( $f'(x) < 0$ ) everywhere it is defined:

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Intervals of Decrease: $(-\infty, -1) \cup (2, \infty)$ because $f'(x) < 0$ .
Intervals of Increase: None.
Conclusion: The function decreases monotonically across both separate valid domain branches and possesses no local extreme points.

Step 4: Inflection Points (Second Derivative Analysis)
Rewrite the derivative as $f'(x) = -3(x^2 - x - 2)^{-1}$ and differentiate using the power chain rule:

$f''(x) = -3 \cdot (-1)(x^2 - x - 2)^{-2} \cdot (2x - 1) = \frac{3(2x - 1)}{(x+1)^2(x-2)^2}$

Setting $f''(x) = 0 \implies 3(2x - 1) = 0 \implies x = \frac{1}{2} = 0.5$ . However, this root falls completely inside the excluded domain gap region where the function does not exist. The squared factors in the denominator are strictly positive where defined, so the curvature sign transitions track along the linear numerator factor $(2x - 1)$ :

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The graph is strictly concave down (concave) on $(-\infty, -1)$ and strictly concave up (convex) on $(2, \infty)$ . Because the mathematical curvature transition occurs across an undefined domain gap, there are no points of inflection.

Step 5: Geometric Curve Sketch

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Sketching an Exponential Function

This is no different from the logarithmic case. In most cases the exponential function should be relatively easy, as a function of the form $e^{ax+b}$ is defined everywhere, so domain and asymptotes should also be easy. Since it is always monotonic, there will be no extreme points. Let us see a slightly more interesting example, even if it may go beyond the scope of the programme.

Example: Conduct a complete analysis for the exponential function $f(x) = e^{\frac{1}{x}}$ . Identify its domain, axes intercepts, horizontal and vertical asymptotes, local extreme points, and points of inflection. Draw a sketch of the curve.

Step 1: Domain and Axes Intercepts
The natural exponential base is defined for any real exponent, but the exponent fraction $\frac{1}{x}$ breaks down when its denominator vanishes at zero.

$\text{Domain: } \mathbb{R} \setminus \{0\} \text{ or } (-\infty, 0) \cup (0, \infty)$

To locate any crossings along the coordinate axes:

$y$ -intercept: 0 is not in the domain of $f$ so there is no $y$ -intercept.
$x$ -intercepts: Setting $f(x) = 0 \implies e^{\frac{1}{x}} = 0$ , which has no real solutions because exponential outputs are strictly positive ( $e^u > 0$ ). The graph has no $x$ -intercepts.

Step 2: Asymptotes

Vertical Asymptotes: Evaluate the one-sided directional limits as approaches the discontinuity marker at zero:
- Approach from the right ( $x \to 0^+$ ): The exponent drives to infinity ( $\frac{1}{x} \to \infty$ ), which forces the output to expand:
  $\lim_{x \to 0^+} e^{\frac{1}{x}} = \infty$
- Approach from the left ( $x \to 0^-$ ): The exponent decays to negative infinity ( $\frac{1}{x} \to -\infty$ ), which creates a finite terminal point:
  $\lim_{x \to 0^-} e^{\frac{1}{x}} = e^{-\infty} = 0$
Because the right-hand limit diverges to infinity, the line $x = 0$ (the -axis) is a vertical asymptote.
Horizontal Asymptotes: Evaluate the limiting behavior on the far left and right tails as $x \to \pm\infty$ :
$\lim_{x \to \pm\infty} \frac{1}{x} = 0 \implies \lim_{x \to \pm\infty} e^{\frac{1}{x}} = e^0 = 1$
This demonstrates that the graph levels off horizontally along the line $y = 1$ .

Step 3: Extreme Points (First Derivative Analysis)
Apply the exponential chain rule ( $\frac{\mathrm{d}}{\mathrm{d}x}[e^u] = u' \cdot e^u$ ) to calculate the first derivative:

$f'(x) = \left(-\frac{1}{x^2}\right)e^{\frac{1}{x}} = -\frac{e^{\frac{1}{x}}}{x^2}$

Setting $f'(x) = 0 \implies -e^{\frac{1}{x}} = 0$ yields no real solutions. Because $e^{\frac{1}{x}} > 0$ and $x^2 > 0$ for all valid values in the domain, the leading negative sign dictates that the derivative remains strictly negative ( $f'(x) < 0$ ):

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Intervals of Decrease: $(-\infty, 0) \cup (0, \infty)$ because $f'(x) < 0$ .
Intervals of Increase: None.
Conclusion: The function decreases continually across both separate tracking branches, leaving it with no local extreme points.

Step 4: Inflection Points (Second Derivative Analysis)
Differentiate $f'(x) = -x^{-2} \cdot e^{x^{-1}}$ using the product rule ( $\left(uv\right)' = u'v + uv'$ ):

$f''(x) = \left(2x^{-3}\right)e^{\frac{1}{x}} + \left(-x^{-2}\right)\left(-x^{-2}e^{\frac{1}{x}}\right) = \frac{2}{x^3}e^{\frac{1}{x}} + \frac{1}{x^4}e^{\frac{1}{x}} = \frac{e^{\frac{1}{x}}(2x + 1)}{x^4}$

Set $f''(x) = 0 \implies 2x + 1 = 0 \implies x = -\frac{1}{2} = -0.5$ . Since $e^{\frac{1}{x}} > 0$ and $x^4 > 0$ where defined, the algebraic sign transitions follow the linear numerator term $(2x + 1)$ :

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The graph actively switches curvature from concave down to concave up across the value $x = -\frac{1}{2}$ . Evaluating the exact matching point height yields:

$f\left(-\frac{1}{2}\right) = e^{\frac{1}{-1/2}} = e^{-2} = \frac{1}{e^2} \approx 0.135$

This confirms a unique point of inflection located at coordinates $\mathbf{\left(-\frac{1}{2}, e^{-2}\right)}$ .

Step 5: Geometric Sketch

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Exercises

For each of the following functions, sketch their graphs. Your solutions should explicitly determine: domain restrictions, axes intercepts, equations of vertical and horizontal asymptotes, local extreme points, and points of inflection.

$f(x) = \ln(2x + 6)$

$f(x) = \ln\left(\frac{x-2}{x+1}\right)$

$f(x) = 3 - e^{-x}$

$f(x) = e^{-\frac{1}{x^2}}$