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
For f(x) = x² + 6x, what is f''(x)?
- 2
- 2x + 6
- 6
- 2x
At what value of x does y = x² - 8x + 3 have a stationary point?
Answer: ______________
Find and classify the stationary point of y = x² - 6x + 11.
- (3, 11) and it is a minimum
- (3, 2) and it is a maximum
- (3, 2) and it is a minimum
- (6, 11) and it is a minimum
Why does a positive second derivative mean a minimum rather than a maximum?
- because the curve itself is positive there
- because the gradient is increasing through the point, so the curve falls then rises
- because a positive number always means the largest value
- because the curve crosses the x-axis at that point
For y = x³ - 6x² + 5, what is f''(2)?
Answer: ______________
Sam solves f'(x) = 0, gets x = 3, and writes the stationary point as (3, 0). Where is the slip?
- the x value should have come from f(x) = 0 instead
- stationary points are always written as a single number
- the y-coordinate comes from the original curve, not from f'(x) = 0
- he should have solved f''(x) = 0 to get the x value
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 is flat there

Answer key

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

Stationary Points and the Second Derivative W1-mt_VbGJEFFgfs-s1

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.
2 · f'(x) = 2x + 6, and differentiating that gives f''(x) = 2. The 6 is a constant in the first derivative, so it goes.
4 · f'(x) = 2x - 8. Setting that to zero gives 2x = 8, so x = 4.
(3, 2) and it is a minimum · f'(x) = 2x - 6 is zero at x = 3. The curve gives y = 9 - 18 + 11 = 2. Then f''(x) = 2, which is positive, so (3, 2) is a minimum.
because the gradient is increasing through the point, so the curve falls then rises · f''(x) reports the trend of the gradient. Positive means the gradient is climbing, so it passes from negative through zero to positive, which is a valley.
0 · f'(x) = 3x² - 12x, so f''(x) = 6x - 12. At x = 2 that is 12 - 12, which is 0.
the y-coordinate comes from the original curve, not from f'(x) = 0 · The 0 he has used is the value of the derivative, not the height. Substituting 3 into the original curve is what gives the y-coordinate.
-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.