A pharmaceutical company is designing a controlled-release drug capsule. The drug concentration in the bloodstream must stay within a therapeutic window: too low and it is ineffective, too high and it is toxic. The concentration follows $C(t) = 5te^{-0.5t}$ mg/L. When is the concentration at its maximum? What is the maximum concentration? These are optimisation problems — the capstone skill of differential calculus. By finding where the derivative equals zero, we can maximise effectiveness while minimising toxicity.
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For $f(x) = xe^{-x}$, do you think the maximum occurs at $x = 0$, $x = 1$, or $x = e$? Predict before calculating.
Core Content
To find stationary points of functions involving exponentials:
Why factor out the exponential? Since $e^{kx} > 0$ for all real $x$, it can never make a product equal to zero. If $f'(x) = e^{kx} \cdot g(x) = 0$, then $g(x) = 0$ is the only equation we need to solve.
Example: Find stationary points of $f(x) = xe^{-x}$.
f'(x) = 1·e^{-x} + x·(-e^{-x}) = e^{-x}(1 - x)
Since $e^{-x} > 0$, solve $1 - x = 0$, giving $x = 1$.
At $x = 1$: $f(1) = 1·e^{-1} = \frac{1}{e} \approx 0.368$.
Second derivative: $f''(x) = -e^{-x}(1-x) + e^{-x}(-1) = e^{-x}(x - 2)$.
At $x = 1$: $f''(1) = e^{-1}(-1) < 0$, so maximum.
Show y = xe^{-x} for x from 0 to 4. Mark the maximum at (1, 1/e ≈ 0.368). Draw the tangent line horizontal at the maximum. Show the curve rising from (0,0) to the maximum, then decaying asymptotically to y=0. Label the x-axis intercept at x=0 and the asymptote y=0.
For word problems:
Find the minimum of $f(x) = x^2 e^x$ for $x < 0$.
The $x$-value and minimum value.
f'(x) = 2x·e^x + x^2·e^x = e^x·x(2 + x)
Set f'(x) = 0: e^x > 0 always, so x(2+x) = 0
x = 0 or x = -2
For x < 0: x = -2 is the critical point
f(-2) = (-2)^2·e^{-2} = 4/e^2 ≈ 0.541
Second derivative test or sign analysis confirms minimum.
Minimum at $x = -2$, value = $\frac{4}{e^2} \approx 0.541$.
Find the maximum of $f(x) = x e^{-2x}$ for $x \geq 0$.
Answer:
$f'(x) = e^{-2x} + x(-2)e^{-2x} = e^{-2x}(1 - 2x) = 0 \Rightarrow x = \frac{1}{2}$. $f(\frac{1}{2}) = \frac{1}{2}e^{-1} = \frac{1}{2e} \approx 0.184$.
First derivative test: Check the sign of $f'(x)$ on either side of the stationary point.
Second derivative test:
For exponential functions, the second derivative test is often efficient because $e^{kx}$ remains positive throughout.
Solve $f'(x) = 0$
$e^{kx} > 0$ always — factor it out
$f''(x) < 0$ or $f'$ changes $+$ to $-$
$f''(x) > 0$ or $f'$ changes $-$ to $+$
The maximum of $f(x) = xe^{-x}$ occurs at $x = 1$ (not $x = 0$ or $x = e$). At $x = 0$, $f(0) = 0$ which is a minimum on the domain $x \geq 0$. At $x = 1$, $f(1) = 1/e \approx 0.368$ which is the maximum. As $x \to \infty$, $e^{-x}$ decays faster than $x$ grows, so $f(x) \to 0$.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Find and classify the stationary point(s) of $f(x) = x^2 e^{-x}$. Show all working. 3 MARKS
9. A drug has concentration $C(t) = 6te^{-0.3t}$ mg/L where $t$ is in hours. Find when the maximum concentration occurs and calculate the maximum concentration to 2 decimal places. 3 MARKS
10. Consider $f(x) = \frac{\ln x}{x}$ for $x > 0$. Find the stationary point, classify it, and determine the behaviour of $f(x)$ as $x \to 0^+$ and as $x \to \infty$. Sketch the curve, showing all key features. 3 MARKS
1. $f'(x) = 2xe^{-x} - x^2e^{-x} = xe^{-x}(2-x)$. Stationary points at $x = 0$ and $x = 2$. $f(0) = 0$ (minimum), $f(2) = 4e^{-2} \approx 0.541$ (maximum).
2. $f'(x) = 3e^{-0.5x}(1 - 0.5x) = 0 \Rightarrow x = 2$. $f(2) = 6e^{-1} = 6/e \approx 2.21$.
3. $f'(x) = e^x + (x+1)e^x = e^x(x+2) = 0 \Rightarrow x = -2$. $f(-2) = (-1)e^{-2} = -1/e^2 \approx -0.135$ (minimum).
1. $e^{-x} > 0$ for all real $x$, so it cannot equal zero. Only the other factor can create zeros.
2. $C'(t) = 8e^{-0.4t}(1 - 0.4t) = 0 \Rightarrow t = 2.5$ hours. $C(2.5) = 20e^{-1} = 20/e \approx 7.36$ mg/L.
3. Maximum at $(1, 1/e)$. Passes through $(0, 0)$. Asymptotic to $y = 0$ as $x \to \infty$.
Q8 (3 marks): $f'(x) = 2xe^{-x} - x^2e^{-x}$ [0.5] $= xe^{-x}(2-x)$ [0.5]. $f'(x) = 0$ when $x = 0$ or $x = 2$ [0.5]. $f''(x) = e^{-x}(2 - 4x + x^2)$ [0.5]. At $x = 0$: $f''(0) = 2 > 0$, so local minimum [0.5]. At $x = 2$: $f''(2) = e^{-2}(2 - 8 + 4) = -2e^{-2} < 0$, so local maximum [0.5].
Q9 (3 marks): $C'(t) = 6e^{-0.3t} + 6t(-0.3)e^{-0.3t}$ [0.5] $= 6e^{-0.3t}(1 - 0.3t)$ [0.5]. $C'(t) = 0$ when $1 - 0.3t = 0$, so $t = \frac{10}{3} \approx 3.33$ hours [0.5]. $C(\frac{10}{3}) = 6 \times \frac{10}{3} \times e^{-1} = 20e^{-1} = \frac{20}{e} \approx 7.36$ mg/L [1.5].
Q10 (3 marks): $f'(x) = \frac{(1/x) \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$ [0.5]. $f'(x) = 0$ when $\ln x = 1$, so $x = e$ [0.5]. $f(e) = \frac{\ln e}{e} = \frac{1}{e} \approx 0.368$ [0.25]. For $x < e$: $\ln x < 1$, so $f'(x) > 0$ (increasing). For $x > e$: $\ln x > 1$, so $f'(x) < 0$ (decreasing). Thus $x = e$ is a local maximum [0.5]. As $x \to 0^+$: $\ln x \to -\infty$, so $f(x) \to -\infty$ [0.25]. As $x \to \infty$: $\ln x$ grows slower than $x$, so $f(x) \to 0^+$ [0.5]. The curve starts from $-\infty$ near $x = 0$, crosses zero at $x = 1$, rises to a maximum at $(e, 1/e)$, then decreases asymptotically to $y = 0$ [0.5].
Climb platforms using stationary points, classification, and exponential optimisation. Pool: lesson 14.
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