Solving dy/dx = 0 gives you possible stationary-point locations, but it does not finish the question. You may also need the coordinates and the nature of each point. Mixing those jobs is how a correct derivative becomes an incomplete answer.
Use a three-part routine: find the x-values, substitute into the original function, then classify the points. The worked example here is original and focuses on a polynomial, so the reasoning is visible without complicated algebra.
Cambridge Mathematics 9709 has different AS and A Level routes. Match your revision to your own components using the official course and syllabus page. This is a focused calculus lesson, not a route-specific syllabus checklist.
Find where the tangent is horizontal
Consider y = x³ − 3x² − 9x + 5.
Differentiate term by term: dy/dx = 3x² − 6x − 9. A stationary point has zero first derivative, so solve 3x² − 6x − 9 = 0.
Divide by 3 and factorise: x² − 2x − 3 = (x − 3)(x + 1) = 0. The candidate x-values are 3 and −1.
It is worth checking this factorisation by expanding. A wrong sign here changes both locations even when the differentiation is correct.
Use the original function for the coordinates
At x = −1, y = −1 − 3 + 9 + 5 = 10. At x = 3, y = 27 − 27 − 27 + 5 = −22. The stationary points are therefore (−1, 10) and (3, −22).
Do not substitute into dy/dx to find the y-coordinate. You already know the derivative is zero at these x-values; that tells you the gradient, not the height of the curve.
Classify with a second derivative
Differentiate again: d²y/dx² = 6x − 6.
At x = −1, the second derivative is −12, so this is a local maximum. At x = 3, it is 12, so this is a local minimum.
For this smooth polynomial, a negative second derivative at a stationary point indicates the curve is locally concave down; a positive value indicates it is locally concave up. These are local statements. The cubic is unbounded, so neither point is an absolute maximum or minimum over all real x.
Know when the test is inconclusive
If the second derivative is zero, do not automatically call the point an inflection point. Compare y = x³ and y = x⁴ at x = 0: both have zero first and second derivatives there, but their shapes differ. The first has a stationary point of inflection; the second has a minimum.
One alternative is to check the sign of the first derivative on either side. A change from positive to negative indicates a local maximum; negative to positive indicates a local minimum. For a constrained interval, check endpoints as well when looking for absolute extrema.
Try a fresh question
Find and classify the stationary point of y = x² − 8x + 7. Write your answer as a coordinate and a type, not just an x-value.
The derivative is 2x − 8, so x = 4. The original function gives y = 16 − 32 + 7 = −9. The second derivative is 2, which is positive, so (4, −9) is a minimum. Completing the square gives y = (x − 4)² − 9 and independently confirms the result.
Continue with StudyForA's Mathematics 9709 lessons. In your next solution, give each line a purpose: gradient, coordinate or classification.