For example, y = 3x3 + 9x2 + 2. Determine the stationary points and their nature.
Let's remind ourselves what a stationary point is, and what is meant by the nature of the points.
A stationary point is a point on a curve where the gradient equals 0.
The nature of a stationary point is:
A minimum - if the stationary point(s) substituded into d2y/dx2 > 0
A point of inflection - if the stationary point(s) substituded into d2y/dx2 = 0 and d2y/dx2 of each side of the point has different signs.
A maximum - if the stationary point(s) substituded into d2y/dx2 < 0
Back to our question...
y = 3x3 + 9x2 + 2. Determine the stationary points and their nature of the curve.
In order to determine the stationary points, we need to differentiate y to get dy/dx. Using standard differentiation...
dy/dx = 9x2 + 18x
We now need to equate dy/dx = 0, as dy/dx = 0 at stationary points.
⇒ 0 = 9x2 + 18x
Solving for x by factorising, we get
⇒ 0 = 3x(3x + 6)
so x = 0 or x = -2.
We have the x values of the stationary points, now we can find the corresponding y values of the points by substituing the x values into the equation for y.
For x = 0,
y = 3(0)3 + 9(0)2 + 2 = 2
So (0, 2) is a stationary point.
For x = -2
y = 3(-2)3 + 9(-2)2 + 2 = 14
So (-2, 14) is a stationary point.
In order to determine the nature of the points, we need to work out d2y/dx2. This means we need to differentiate our dy/dx. By using standard differentiation...
dy/dx = 9x2 + 18x
⇒ d2y/dx2 = 18x + 18
We will now substitute both of our stationary points into d2y/dx2 to determine their nature.
For x = 0
d2y/dx2 = 18
As 18 > 0, this point is a minimum.
For x = -2,
d2y/dx2 = -18.
As -18 < 0, this point is a maximum.
Our answer is: Stationary point 1 is (0, 2) - a minimum, and stationary point 2 is (-2, 14), a maximum.
Here are a few more questions to test your understanding, scroll down for the answers!
Find the stationary points on the curve y = 3x2 - 12x, and determine their nature.
Find the stationary points on the curve y = x2ex, and determine their nature.
Answers
The stationary point is (-2, -12), a minimum
The stationary points are (0,0), a minimum, and (-2, 0.541), a maximum.