Stationary Points and the Second Derivative

Mathematics · Calculus & Analysis · ages 17-18

Name ______________________ Date ____________

If f''(x) is negative at a stationary point, that point is a maximum.
Circle one: True False
To get the y-coordinate of a stationary point, substitute the x value into f'(x).
Circle one: True False
The curve y = 5 - x² has one stationary point. What kind is it?
- a point of inflection
- a minimum
- the test cannot say
- a maximum
For f(x) = x² + 6x, what is f''(x)?
- 2
- 2x + 6
- 6
- 2x
For y = 12x - x³, what kind of stationary point sits at x = 2?
- a minimum
- a point of inflection
- the test cannot say
- a maximum
For y = x³ - 12x, what kind of stationary point sits at x = 2?
- a maximum
- a minimum
- a point of inflection
- the test cannot say
If f''(x) = 0 at a stationary point, that point has to be a point of inflection.
Circle one: True False
For y = x³ - 6x² + 9x + 1, what is the value of f''(1)?
Answer: ______________
The curve y = x³ + 3x² - 9x has a minimum. What is its y-coordinate?
Answer: ______________
Ana finds a stationary point where f''(x) = 0 and concludes that it is neither a maximum nor a minimum. Where is the slip?
- a zero second derivative settles nothing, so check the sign of f'(x) either side
- f''(x) can never be zero at a stationary point
- she should have concluded it is a minimum, since zero is not negative
- she should have said it is a point of inflection, since the curve looks flat there

Answer key

For grown-ups. Fold this page away before handing over the rest.

True · A negative second derivative means the gradient is falling through the point, so it goes from positive to zero to negative. That is a peak.
False · f'(x) is zero at a stationary point by definition, so it would always give 0. Heights come from the original curve.
a maximum · f'(x) = -2x, which is zero at x = 0. Then f''(x) = -2, which is negative, so the point is a maximum.
2 · f'(x) = 2x + 6, and differentiating that gives f''(x) = 2. The 6 is a constant in the first derivative, so it goes.
a maximum · f'(x) = 12 - 3x², so f''(x) = -6x. At x = 2 that is -12, which is negative, so the point is a maximum.
a minimum · f'(x) = 3x² - 12, so f''(x) = 6x. At x = 2 that is 12, which is positive, so the point is a minimum.
False · A zero second derivative settles nothing. For y = x⁴ both derivatives are zero at x = 0, and that point is a minimum, not an inflection.
-6 · f'(x) = 3x² - 12x + 9 and f''(x) = 6x - 12, so f''(1) = 6 - 12 = -6.
-5 · f'(x) = 3x² + 6x - 9 = 3(x + 3)(x - 1), so the stationary points are at x = -3 and x = 1. f''(x) = 6x + 6 is 12 at x = 1, positive, so that is the minimum, and the curve gives 1 + 3 - 9 = -5.
a zero second derivative settles nothing, so check the sign of f'(x) either side · Zero is not a third verdict, it is the absence of one. y = x⁴ has both derivatives zero at x = 0 and still has a minimum there.