Differentiation

Have a go first. Then open the working one line at a time.

1. Turning point

   Halfway between −2 and 4: x = 1

   f is a parabola, so its turning point is halfway between its roots. The gradient there is 0, so f' has a root at x = 1.

2. Signs of the gradient

   f falls for x < 1, rises for x > 1

   So f' is negative to the left of 1 and positive to the right. A parabola has one turning point, so f' is a straight line.

3. The sketch

   f'(x) = 2x − 2: a straight line through (1, 0) with positive gradient

   f' has degree one less than f. Mark the root at 1 and the y-intercept at −2.

Have a go first. Then open the working one line at a time.

1. Turning points

   f has a maximum between −2 and 1 and a minimum between 1 and 3

   A cubic with three roots has two turning points. Each one is a root of f'.

2. From the calculator

   Graph f and find the turning points: x = −0.786 and x = 2.12

   Read them from the graph, to 3 significant figures.

3. Shape of f'

   f' is a parabola, positive for x < −0.786 and for x > 2.12, negative between

   f' has degree two. f rises, falls, then rises again, so f' is positive, negative, positive: an upward parabola.

Have a go first. Then open the working one line at a time.

1. Sign of f'

   f is decreasing everywhere, so f'(x) < 0 for all x

   The whole graph of f' lies below the x-axis.

2. Far to the right

   f levels off along y = 1, so f'(x) → 0

   A horizontal asymptote means the gradient tends to 0. The graph of f' has the x-axis as its asymptote.

3. Far to the left

   f is steeper as x decreases, so f'(x) becomes more negative

   f' is below the axis, increasing towards 0 as x increases.

Have a go first. Then open the working one line at a time.

1. Stationary points

   f'(−1) = 0 and f'(3) = 0

   f has a turning point at each root of f'.

2. Max or min?

   f' > 0 for x < −1, f' < 0 for −1 < x < 3, f' > 0 for x > 3

   f rises, falls, then rises: a maximum at x = −1 and a minimum at x = 3.

3. Two possible sketches

   f(x) = x³/3 − x² − 3x + c, for example c = 0 and c = 4

   A cubic with that shape. Moving the graph up or down does not change its gradient, so any c works.

Have a go first. Then open the working one line at a time.

1. (a) Sketch g'

   g rises from 0 to a maximum at x = π/2, so g' starts at 2 and falls through 0 at x = π/2

   The gradient of a sine wave follows a cosine wave: g'(x) = 2 cos x, from 2 down to −2 and back up to 2.

2. (b) A possible f

   f' = 2 sin x, so f(x) = −2 cos x + c

   f' is positive on (0, π), so f rises to a maximum at x = π. Then f' is negative, so f falls.

3. Not unique

   f(x) = −2 cos x + c for any constant c

   Adding c moves the graph up or down. It does not change any gradient, so every c gives the same f'.

4. (c) Maximum

   f(0) = −2 + c = 5, so c = 7. Maximum at x = π: f(π) = 2 + 7 = 9

   f is largest where f' goes from positive to negative, at x = π, where cos π = −1.

Have a go first. Then open the working one line at a time.

1. Sketch

   From the calculator: graph \(y=x^2(x^2-5)\), a W-shape that touches the axis at x = 0

   The factor x² gives a repeated root at 0: the curve touches the axis there.

2. (a) f(x) > 0

   Roots of f: x = −2.24, 0, 2.24. f(x) > 0 for x < −2.24 and x > 2.24

   Read where the curve is above the x-axis. At x = 0 the curve touches the axis, so f(0) = 0 is not greater than 0.

3. Turning points

   From the calculator: turning points at x = −1.58, 0 and 1.58

   Each turning point is a root of f'.

4. (b) f'(x) < 0

   f falls for x < −1.58 and for 0 < x < 1.58

   Where the curve goes down as you move right, the gradient is negative.

Have a go first. Then open the working one line at a time.

1. Turning points of A

   A has a minimum at x ≈ −1.73 and a maximum at x ≈ 1.73

   The derivative graph must have roots at both. B and C both do.

2. Compare the signs

   A rises between −1.73 and 1.73, so f' > 0 there

   B is positive between its roots. C is negative between its roots. So B is f'.

3. Check the degree

   A is a cubic, B is a parabola

   The derivative of a cubic is a quadratic: one degree less.

4. Local maximum

   f'(x) = 3 − x² = 0, so x = √3 = 1.73

   f' changes from positive to negative at x = 1.73: a maximum.