Increasing/decreasing functions; sign of f'(x) practice questions.

Which condition on f′(x) throughout an interval shows that f is increasing on that interval?

• f′(x) < 0
• f′(x) > 0
• f′(x) = 0
• f′(x) is undefined

Answer: f′(x) > 0

A function is increasing where its gradient is positive, i.e. f′(x) > 0.

For f(x) = x³ − 3x, determine the interval on which f is decreasing.

• x < −1
• −1 < x < 1
• x > 1
• all real x

Answer: −1 < x < 1

f′(x) = 3x² − 3 = 3(x−1)(x+1); this is negative between the roots, i.e. for −1 < x < 1.

At a local maximum of a smooth curve, determine what happens to the sign of f′(x) as x increases through the maximum.

• It changes from negative to positive
• It changes from positive to negative
• It stays positive throughout
• It stays negative throughout

Answer: It changes from positive to negative

Before a maximum the curve rises (f′ > 0) and after it falls (f′ < 0), so f′ changes from positive to negative.

The gradient function of f is f′(x) = (x − 2)(x + 3). Determine the values of x for which f is increasing.

• −3 < x < 2
• x < −3 or x > 2
• x < 2 only
• x > −3 only

Answer: x < −3 or x > 2

The quadratic (x−2)(x+3) opens upward and is positive outside its roots, so f′(x) > 0 (increasing) for x < −3 or x > 2.

For f(x) = x² + 2x + 7, determine whether f is increasing or decreasing at x = 0.

• Increasing, because f′(0) = 2 > 0
• Decreasing, because f′(0) = −2 < 0
• Stationary, because f′(0) = 0
• Increasing, because f(0) = 7 > 0

Answer: Increasing, because f′(0) = 2 > 0

f′(x) = 2x + 2, so f′(0) = 2 > 0, meaning f is increasing at x = 0.

A profit function is P(x) = −2x² + 40x − 50 (in $1000s), where x is thousands of units sold. Determine the number of units (in thousands) that maximises profit.

• x = 10
• x = 20
• x = 40
• x = 5

Answer: x = 10

P′(x) = −4x + 40 = 0 gives x = 10; the negative leading coefficient confirms a maximum.