Introduction
Exponential and logarithmic functions can appear alone or inside more complicated expressions. Their derivatives often have factors that are always positive, which makes the sign analysis efficient. The key is to identify the domain, find the critical numbers that lie in it, and determine the derivative's sign on either side of each one.
A critical number is a candidate, not automatically an extremum. The derivative may change from positive to negative, from negative to positive, or not change sign at all. A sign chart distinguishes these possibilities and explains how the function behaves near the candidate.
Domains and Critical Numbers
The exponential function \(e^x\) is defined and positive for every real \(x\). The natural logarithm \(\ln u\) is defined only when its input \(u\) is positive. For a function involving a logarithm, first solve the inequality that makes its input positive; that condition determines where the function exists.
For example, if \(f(x)=\ln(x-2)\), its input must satisfy \(x-2>0\), so the domain is \(x>2\). A number outside that domain cannot be a critical number of this function, even if an algebraic derivative expression appears to have a zero there.
In this tutorial, local maxima and minima are classified using derivative signs. At an interior point \(c\), a change from positive to negative means the function rises and then falls, giving a local maximum. A change from negative to positive means it falls and then rises, giving a local minimum. No sign change means neither conclusion follows from the first derivative sign test.
Why Derivative Sign Changes Classify Extrema
The sign test follows from the Mean Value Theorem. Suppose \(f\) is continuous near \(c\), differentiable on each side of \(c\), and \(f'(x)>0\) just to the left of \(c\), while \(f'(x)<0\) just to its right. For any nearby \(x<c\), the Mean Value Theorem applied on \([x,c]\) gives a number \(\xi\) between \(x\) and \(c\) such that
Thus \(f(x)<f(c)\) to the left. For any nearby \(x>c\), applying the theorem on \([c,x]\) gives a number \(\eta\) between \(c\) and \(x\) such that
Thus \(f(x)<f(c)\) on the right as well, so \(f(c)\) is a local maximum. Reversing both derivative signs gives \(f(x)>f(c)\) on both sides, proving a local minimum. This argument explains why the sign change, rather than merely the equation \(f'(c)=0\), determines the classification.
Worked Example: An Exponential Function
Worked Example: An Exponential Function
Find and classify the local extrema of \(f(x)=(x-1)e^{-x}\).
Step 1: Check the domain. The exponential factor is defined for every real \(x\), so the domain is all real numbers.
Step 2: Differentiate. Apply the product rule and use \(\frac{d}{dx}e^{-x}=-e^{-x}\):
Step 3: Find critical numbers and signs. The factor \(e^{-x}\) is positive for all \(x\). Therefore \(f'(x)=0\) exactly when \(2-x=0\), giving the critical number \(x=2\). For \(x<2\), \(2-x>0\), so \(f'(x)>0\). For \(x>2\), \(2-x<0\), so \(f'(x)<0\). The derivative changes from positive to negative, so \(x=2\) is a local maximum.
Step 4: Find the function value and check.
Thus the local maximum occurs at \((2,e^{-2})\). The derivative formula is consistent with the classification: \(f'(2)=e^{-2}(2-2)=0\), and its sign is positive immediately to the left and negative immediately to the right.
Worked Example: A Logarithmic Function
Worked Example: A Logarithmic Function
Find and classify the local extrema of \(g(x)=\frac{\ln x}{x}\).
Step 1: Check the domain. The logarithm requires \(x>0\), so the domain is \((0,\infty)\).
Step 2: Differentiate. Using the quotient rule and \(\frac{d}{dx}\ln x=\frac{1}{x}\),
Step 3: Find critical numbers and signs. In the domain, \(x^2>0\), so the sign of \(g'(x)\) is determined by \(1-\ln x\). Setting the numerator equal to zero gives \(\ln x=1\), hence \(x=e\). For \(0<x<e\), the increasing property of \(\ln x\) gives \(\ln x<1\), so \(g'(x)>0\). For \(x>e\), \(\ln x>1\), so \(g'(x)<0\). Therefore \(x=e\) is a local maximum.
Step 4: Find the function value and check.
The local maximum is \((e,1/e)\). Substituting \(x=e\) into the derivative gives \(g'(e)=(1-\ln e)/e^2=0\), as required for this interior critical number.
Exponential and Logarithmic Factors in Compositions
For a composite function, the chain rule can produce a numerator whose sign changes while the remaining factors stay positive. For instance, if \(F(x)=\ln(u(x))\), the function requires \(u(x)>0\), and its derivative is \(F'(x)=u'(x)/u(x)\). On the domain, the denominator \(u(x)\) is positive, so the sign of the derivative is the sign of \(u'(x)\).
Similarly, differentiating \(e^{v(x)}\) gives \(e^{v(x)}v'(x)\). Since \(e^{v(x)}>0\) wherever it is defined, its derivative has the sign of \(v'(x)\). These observations can simplify a sign chart, but they do not remove the need to check the domain or the remaining factors.
Worked Example: A Logarithm of Exponential Terms
Worked Example: A Logarithm of Exponential Terms
Find and classify the local extrema of \(h(x)=\ln(e^x+4e^{-x})\).
Step 1: Check the domain. Both exponential terms are positive for every real \(x\), so their sum is positive. The logarithm is therefore defined for all real \(x\).
Step 2: Differentiate using the chain rule. The derivative of \(e^x+4e^{-x}\) is \(e^x-4e^{-x}\), giving
The denominator is positive everywhere, so the sign of \(h'(x)\) is determined by its numerator. Multiply that numerator by the positive quantity \(e^x\) to make its sign easier to read:
Step 3: Find the critical number and signs. The numerator is zero when \(e^{2x}=4\). Since \(e^x>0\), this is equivalent to \(e^x=2\), so \(x=\ln 2\). If \(x<\ln 2\), then \(e^x<2\), hence \(e^{2x}<4\) and \(h'(x)<0\). If \(x>\ln 2\), then \(e^x>2\), hence \(e^{2x}>4\) and \(h'(x)>0\). The derivative changes from negative to positive, so \(x=\ln 2\) is a local minimum.
Step 4: Find the function value and check.
The local minimum is \((\ln 2,\ln 4)\). Direct substitution in the derivative confirms \(h'(\ln 2)=(2-2)/(2+2)=0\), and the sign analysis confirms that the function decreases before this point and increases after it.
A Reliable Sign-Analysis Process
For an exponential or logarithmic function, organize the work so that domain restrictions and signs are explicit. A useful sequence is:
For every logarithm, require its input to be positive. Exponential expressions alone accept every real input.
Apply the product, quotient, or chain rule as needed, and simplify without discarding factors that could affect the sign.
Solve \(f'(x)=0\) and note where the derivative is undefined while the function itself is defined.
Use factors known to be positive to isolate the part of the derivative that determines its sign.
A positive-to-negative change gives a local maximum; a negative-to-positive change gives a local minimum. Substitute the critical number into the original function to report the extremum value.
Common Pitfall: Treating Every Critical Number as an Extremum
A zero derivative does not by itself prove that there is a local maximum or minimum. For example, consider \(p(x)=e^x\). Its derivative is \(p'(x)=e^x\), which is positive for every real \(x\). There are no critical numbers and no local extrema. More generally, a critical number can have a derivative that touches zero without changing sign. That is why the sign on both sides must be checked.
Another error is to ignore the domain when solving for a critical number. For \(q(x)=\ln x\), the equation \(q'(x)=1/x=0\) has no solution, and \(x=0\) is not a critical number: it is not in the domain. A derivative expression's algebraic features do not make a point part of the function's domain.
Check Your Understanding
For each function, determine its domain, find any critical numbers, and use derivative signs to classify local extrema.
- For \(f(x)=(x+1)e^{-x}\), find the critical number and the local extremum value.
- For \(g(x)=\ln x-x\), find and classify every critical number in its domain.
- For \(h(x)=\ln(e^x+e^{-x})\), locate its critical number and determine whether it is a local maximum or minimum.
- Why does a factor \(e^{v(x)}\) not change the sign of \(v'(x)\) in the derivative of \(e^{v(x)}\)?
- What derivative sign change indicates a local minimum?
