Use differentiation to find stationary points of functions

What is a stationary point?

A point on the graph of a function where the slope of the tangent line is zero. In other words, the derivative $f'(x)$ equals zero.

Analogy: Imagine walking on a hill. A stationary point is like reaching the top of a hill (maximum) or the bottom of a valley (minimum) – you’re not going up or down at that instant.

Finding stationary points: Step‑by‑step

1. Write down the function $f(x)$.
2. Differentiate to get $f'(x)$.
3. Set the derivative equal to zero: $f'(x)=0$.
4. Solve the equation for $x$ (these are the critical points).
5. Plug each critical point back into the original function to find the y‑coordinate.
6. Use the second derivative test or the first derivative test to classify each point as a maximum, minimum, or saddle point.

Example: Find stationary points of $f(x)=x^3-3x+2$

1️⃣ Differentiate:

$f'(x)=3x^2-3$

2️⃣ Set derivative to zero:

$3x^2-3=0 \;\Rightarrow\; x^2=1 \;\Rightarrow\; x=\pm1$

3️⃣ Find y‑values:

4️⃣ Second derivative test:

$f''(x)=6x$

• At $x=1$: $f''(1)=6>0$ → local minimum at $(1,0)$.

• At $x=-1$: $f''(-1)=-6<0$ → local maximum at $(-1,4)$.

Exam Tips 📌

• Always show the full derivative calculation – examiners look for clear steps.
• When solving $f'(x)=0$, check for extraneous solutions (e.g., division by zero).
• Use the second derivative test whenever possible; it’s faster than the first derivative test.
• Remember to include both the x‑ and y‑coordinates of each stationary point.
• Check your final answer against the domain of the function (e.g., avoid points where the function isn’t defined).

Quick Practice Question

Find the stationary points of $g(x)=\frac{1}{x^2}+4x$ and classify them.

🧠 Hint: First differentiate, then solve $g'(x)=0$ and apply the second derivative test.