Find the minimum and maximum values of f(x) = e^x-ax
Find the minimum and maximum values of $f(x) = e^{x}-ax$, where $a\gt 0$.
Explanation of Method
One critical point, and it is the global minimum.
Step 1: The derivative and the critical point
The function is defined on all of $\mathbb{R}$, and
\[ f'(x) = e^{x}-a = 0 \;\Longrightarrow\; x = \ln a, \tag{1} \]
which exists precisely because $a\gt 0$.
Step 2: Sign of the derivative
Since $e^{x}$ is increasing, $e^{x}\lt a$ for $x\lt \ln a$ and $e^{x}\gt a$ for $x\gt \ln a$: $x\lt \ln a:\ f'\lt 0\ (\searrow); \qquad x\gt \ln a:\ f'\gt 0\ (\nearrow)$.
The single change from $-$ to $+$ marks a minimum, and since it is the only critical point the minimum is global.
Step 3: Behaviour at infinity
\[ x\to+\infty:\ e^{x} \text{ outgrows } ax \;\Longrightarrow\; f\to+\infty; \qquad x\to-\infty:\ e^{x}\to0 \text{ and } -ax\to+\infty \;\Longrightarrow\; f\to+\infty. \]
The function is unbounded above at both ends, so it has no maximum.
Step 4: The minimum value
\[ f\bigl(\ln a\bigr) = e^{\ln a}-a\ln a = a-a\ln a = a\bigl(1-\ln a\bigr). \tag{2} \]
Step 5: Verification
\[f'' = e^{x}\gt 0 \text{ everywhere: } f \text{ is convex, so the critical point is the global minimum} \;\;\checkmark\]
\[a = 2:\ \min = 2\bigl(1-\ln2\bigr) = 0.6137056 \text{ at } x = \ln2 = 0.6931472 \;\;\checkmark\]
\[a = e:\ \min = e\bigl(1-1\bigr) = 0,\ \text{the line } y = ex \text{ being tangent to } y = e^{x} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\min_{\mathbb{R}}f = f\bigl(\ln a\bigr) = a\bigl(1-\ln a\bigr); \qquad \text{no maximum}\,}\]
Remark
Remark. The sign of the minimum tells when the line $y = ax$ meets the curve $y = e^{x}$: the minimum $a(1-\ln a)$ is positive for $a\lt e$ (no intersection), zero for $a = e$ (tangency), and negative for $a\gt e$ (two intersections). The inequality $e^{x}\ge ax+a(1-\ln a)$, which the result expresses, is the convexity of the exponential written out.