Stationary points; local maxima and minima practice questions.

The function f(x) = x² − 6x + 5 has one stationary point. Determine its x-coordinate.

• x = 3
• x = 6
• x = −3
• x = 5

Answer: x = 3

f′(x) = 2x − 6 = 0 gives x = 3, the vertex of the parabola.

For f(x) = x³ − 3x, determine the nature of the stationary point at x = 1.

• Local minimum
• Local maximum
• Point of inflexion
• No stationary point there

Answer: Local minimum

f′(x) = 3x² − 3 = 0 at x = ±1; f″(x) = 6x, and f″(1) = 6 > 0, so x = 1 is a local minimum.

The curve f(x) = 2x³ − 9x² + 12x + 1 has two stationary points. Determine their x-coordinates.

• x = 1 and x = 2
• x = 0 and x = 3
• x = −1 and x = −2
• x = 2 and x = 3

Answer: x = 1 and x = 2

f′(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2), giving x = 1 and x = 2.

A local minimum of f(x) = x² − 8x + 20 occurs at its stationary point. Calculate the minimum value of f(x).

• 4
• 20
• −4
• 16

Answer: 4

f′(x) = 2x − 8 = 0 gives x = 4; f(4) = 16 − 32 + 20 = 4.

For a function f, the second derivative test states that at a stationary point x = a where f′(a) = 0, if f″(a) < 0 then x = a is which type of point?

• Local maximum
• Local minimum
• Point of inflexion
• A discontinuity

Answer: Local maximum

A negative second derivative means the curve is concave down, so the stationary point is a local maximum.

An open box has volume modelled by V(x) = 60x − 4x² cm³, where x is a side length in cm. Determine the value of x that maximises the volume.

• 7.5
• 15
• 30
• 3.75

Answer: 7.5

V′(x) = 60 − 8x = 0 gives x = 7.5; V″(x) = −8 < 0 confirms a maximum.