Introduction
Exponential and logarithmic functions have distinctive global shapes. An exponential curve has a horizontal asymptote at one end of its domain, while a logarithmic curve has a vertical asymptote at the edge of its domain. Shifts, reflections, and scaling move or alter these features, but the formulas still give a systematic way to locate them.
A reliable sketch combines several kinds of information: the domain, intercepts, limits, the sign of the first derivative, and the sign of the second derivative. No one feature determines the entire graph. Together, they describe where the curve exists, how it moves, and how it bends.
Features of Exponential and Logarithmic Functions
For \(b>0\) with \(b\ne 1\), the exponential function \(b^x\) is defined for every real \(x\), and \(b^x>0\). Its derivative is \((\ln b)b^x\), and its second derivative is \((\ln b)^2b^x\). Thus \(b^x\) is concave up everywhere. It increases if \(b>1\) and decreases if \(0<b<1\).
The natural logarithm \(\ln x\) is defined only for \(x>0\). Its derivative is \(1/x\), and its second derivative is \(-1/x^2\). Therefore \(\ln x\) is increasing and concave down throughout its domain. It crosses the horizontal axis at \(x=1\), because \(\ln 1=0\), and it approaches \(-\infty\) as \(x\) approaches zero from the right.
For example, the shifted and scaled exponential \(f(x)=A e^{k(x-h)}+d\), with \(A\ne 0\) and \(k\ne 0\), has domain \(\mathbb{R}\). Its derivatives are
Since the exponential factor is positive, the sign of \(f'\) is the sign of \(Ak\), and the sign of \(f''\) is the sign of \(A\). The curve approaches the horizontal asymptote \(y=d\) at the end where \(k(x-h)\to-\infty\). At the other end, the exponential term grows in magnitude.
A shifted and scaled logarithm \(g(x)=A\ln(k(x-h))+d\), where \(A\ne 0\) and \(k\ne 0\), is defined only when \(k(x-h)>0\). On that domain,
The boundary \(x=h\) is a vertical asymptote: as the argument \(k(x-h)\) tends to zero from above, its logarithm tends to \(-\infty\). The sign of \(A\) determines whether the function itself tends to \(-\infty\) or \(+\infty\) there. The argument must be positive; it is not enough to notice that \(x-h\) appears in the formula.
A Systematic Sketching Method
For a particular exponential or logarithmic function, work through the following features before drawing the curve.
Exponential expressions are defined for every real input. For a logarithm, solve the strict inequality that its argument must be positive.
Evaluate \(f(0)\) if zero is in the domain. For an \(x\)-intercept, solve \(f(x)=0\), checking that the resulting input belongs to the domain.
Find the limits at infinite ends of the domain and, for logarithms, at finite domain boundaries. This identifies asymptotes and the direction in which the curve escapes.
Use \(f'\) to determine where the function increases or decreases, and \(f''\) to determine where it is concave up or down.
Mark exact intercepts and asymptotes, then connect the features without contradicting the domain, limits, or derivative signs.
The method works because the derivative signs describe local changes across intervals, while limits describe behavior at the edges of those intervals. Concavity adds information about how the slope changes. A sketch that satisfies all three is much more constrained than a curve drawn from a few sample points alone.
Worked Examples
Worked Example: A Basic Exponential Curve
Sketch \(f(x)=2^x\). The exponential is defined for all real inputs, so its domain is \(\mathbb{R}\). Since \(2^x>0\), it has no \(x\)-intercept. At \(x=0\), \(f(0)=2^0=1\), so the \(y\)-intercept is \((0,1)\).
Its derivatives are
Because \(\ln 2>0\) and \(2^x>0\), both derivatives are positive everywhere. The function is increasing and concave up. Its limits are
Thus \(y=0\) is a horizontal asymptote as \(x\to-\infty\). The sketch rises through \((0,1)\), staying above the horizontal axis. Check: substituting \(x=0\) gives \(2^0=1\), and the positive formula for \(f'\) confirms the stated increase.
Worked Example: A Shifted Exponential with an \(x\)-Intercept
Analyze \(f(x)=3-2e^{-x}\). Its domain is \(\mathbb{R}\). At \(x=0\), \(f(0)=3-2=1\), so its \(y\)-intercept is \((0,1)\). To find its \(x\)-intercept, solve
The intercept is therefore \(\bigl(\ln(2/3),0\bigr)\). The derivatives are \(f'(x)=2e^{-x}>0\) and \(f''(x)=-2e^{-x}<0\), so the curve is increasing and concave down everywhere. For the limits, \(e^{-x}\to 0\) as \(x\to\infty\), while \(e^{-x}\to\infty\) as \(x\to-\infty\). Hence
The horizontal asymptote is \(y=3\) as \(x\to\infty\); the curve approaches it from below because \(f(x)=3-2e^{-x}<3\). Check the intercept by substituting \(x=\ln(2/3)\): \(e^{-x}=3/2\), so \(f(x)=3-2(3/2)=0\).
Worked Example: A Shifted Logarithmic Curve
Analyze \(g(x)=\ln(x-1)-1\). Its argument must be positive, so \(x-1>0\), giving the domain \(x>1\). There is no \(y\)-intercept because \(x=0\) is outside this domain. For the \(x\)-intercept,
The derivatives on the domain are \(g'(x)=1/(x-1)>0\) and \(g''(x)=-1/(x-1)^2<0\). Thus the curve increases and is concave down. At the left boundary, \(\ln(x-1)\to-\infty\) as \(x\to1^+\), so \(x=1\) is a vertical asymptote. As \(x\to\infty\), \(g(x)\to\infty\). The sketch rises from \(-\infty\) near \(x=1\), crosses at \(1+e\), and continues upward with decreasing slope. Check the intercept: substituting \(x=1+e\) gives \(\ln(e)-1=1-1=0\).
Worked Example: A Reflected Logarithm
Analyze \(h(x)=1-2\ln(3-x)\). Its argument condition is \(3-x>0\), so the domain is \(x<3\). The \(y\)-intercept is \(h(0)=1-2\ln 3\). To find the \(x\)-intercept, solve
Differentiating gives \(h'(x)=2/(3-x)>0\) and \(h''(x)=2/(3-x)^2>0\) for every \(x<3\). The curve is increasing and concave up. As \(x\to3^-\), \(\ln(3-x)\to-\infty\), so \(h(x)\to\infty\); therefore \(x=3\) is a vertical asymptote. As \(x\to-\infty\), \(\ln(3-x)\to\infty\), so \(h(x)\to-\infty\). Check the root: at \(x=3-\sqrt e\), the argument is \(\sqrt e\), and \(1-2\ln(\sqrt e)=1-2(1/2)=0\).
How Derivative Signs Support the Sketch
The sign tests used in these examples follow directly from the first- and second-derivative tests for shape. On an interval where \(f'(x)>0\), the function increases; where \(f'(x)<0\), it decreases. Where \(f''(x)>0\), the graph is concave up; where \(f''(x)<0\), it is concave down. For the exponential and logarithmic families here, the derivative formulas have fixed signs on each domain, so the curve has the same direction and concavity throughout that domain.
For instance, \(e^u>0\) for every real \(u\), and \((x-h)^2>0\) whenever \(x\ne h\). These facts explain why the signs of the derivatives of transformed functions are controlled by their coefficients, not by the exponential or squared denominator factors. The domain still matters: a derivative formula for a logarithm describes behavior only where its logarithm is defined.
Common Pitfall: Ignoring the Logarithm's Argument
A common error is to assume that \(\ln(3-x)\) is defined for \(x>3\), because the expression contains \(x\). That reverses the domain inequality. The correct condition is \(3-x>0\). Subtracting \(3\) gives \(-x>-3\), and dividing by \(-1\) reverses the inequality, so \(x<3\).
There is a related derivative error. Writing \(\frac{d}{dx}\ln(3-x)=1/(3-x)\) omits the derivative of the inside expression. By the chain rule,
The missing negative sign would incorrectly reverse the predicted direction of the curve. Checking both the argument condition and the chain rule prevents this mistake.
Why These Features Matter
A sketch is not just a drawing that passes through a few points. Intercepts anchor the curve at specific locations; limits identify its boundary behavior; derivative signs determine its direction; and concavity constrains how its slope changes. These checks are especially useful when a graph has been shifted or reflected, since a familiar parent curve may no longer have its usual orientation or intercepts.
Before accepting a sketch, verify that it respects the domain, that each claimed intercept satisfies the original equation, and that its ends agree with the calculated limits. Then check that the direction and bending agree with the derivative signs. Any disagreement signals that the sketch or an earlier calculation needs correction.
Check Your Understanding
For each question, use the domain, limits, and derivative information to justify your answer.
- For \(f(x)=4^x\), state the domain, intercepts, monotonicity, concavity, and horizontal asymptote.
- For \(g(x)=2+e^{x}\), what horizontal asymptote does the curve approach as \(x\to-\infty\), and from which side?
- Find the domain and vertical asymptote of \(h(x)=\ln(5-x)\).
- For \(p(x)=-\ln(x+2)\), determine whether it is increasing or decreasing and whether it is concave up or down on its domain.
- Explain why solving a logarithmic equation for an intercept must be followed by a check that the resulting input lies in the function's domain.
