Sketching f and f' from graphs Problem set B One question at a time

The graph shows \(y=f(x)\), a parabola with roots at \(x=-4\) and \(x=2\). Sketch \(y=f'(x)\).

• Turning point halfway between the roots: x = −1, so f'(−1) = 0
• f falls then rises, so f' is negative then positive
• f'(x) = 2x + 2: a line through (−1, 0) and (0, 2)

The graph shows \(y=f(x)\), a parabola that touches the x-axis at \(x=-2\). Sketch \(y=f'(x)\).

• f touches the axis at x = −2: a turning point, so f'(−2) = 0
• f rises for x < −2 and falls for x > −2, so f' is positive then negative
• f'(x) = −2x − 4: a line through (−2, 0) and (0, −4)

The graph shows \(y=f(x)\), a cubic with roots at \(x=-3\), \(1\) and \(4\). Sketch \(y=f'(x)\), giving the x-values where it crosses the x-axis.

• f has a minimum between −3 and 1 and a maximum between 1 and 4
• From the calculator: turning points at x = −1.36 and x = 2.69
• f' is negative, positive, negative: a downward parabola with those roots
• f'(0) = 11, the gradient of f at x = 0

The graph shows \(y=f(x)\), a cubic that touches the x-axis at \(x=2\) and crosses it at \(x=-1\). Sketch \(y=f'(x)\).

• The touching point x = 2 is a turning point, so f'(2) = 0
• The other turning point is the maximum between −1 and 2, at x = 0 (f(0) = 4)
• f' is positive, negative, positive: upward parabola with roots 0 and 2

The graph shows \(y=f(x)\), a decreasing curve with the horizontal asymptote \(y=-1\). It crosses the x-axis at \(x=1.39\). Sketch \(y=f'(x)\).

• f is decreasing for all x, so f'(x) < 0: the graph is below the x-axis
• The asymptote of f means f'(x) → 0 for large x
• f gets steeper to the left, so f' is more negative there: f'(0) = −4

Sketch a possible graph of a function f that satisfies each condition.

1. (a) \(f(x)<0\) for all x, and \(f'(x)<0\) for all x.

   • f < 0 means the curve is below the x-axis
   • f' < 0 means it falls everywhere: −e^x does both

2. (b) f is always decreasing, and \(f'\) is always increasing.

   • f always decreasing: f' < 0
   • f' always increasing: the gradient moves up towards 0, so the curve flattens but keeps falling

The graph shows \(T'(t)\), the rate of change of the temperature of a room in °C per hour, for \(0\le t\le 6\). It is a parabola with roots at \(t=1\) and \(t=5\). Sketch two possible graphs of the temperature \(T(t)\).

• T' > 0 for t < 1: T rises
• T' < 0 for 1 < t < 5: T falls, with a maximum at t = 1 and a minimum at t = 5
• T' > 0 for t > 5: T rises again
• Both curves have the same shape, one shifted up: T(t) = t³/3 − 3t² + 5t + c, for example c = 15 and c = 18

The graph shows \(y=f(x)=2\sin x\) for \(0\le x\le 2\pi\).

1. (a) Sketch \(y=f'(x)\).

   • f has a maximum at π/2 and a minimum at 3π/2
   • So f' = 0 at those points, with f'(0) = 2 the steepest rise
   • f'(x) = 2 cos x

2. (b) Taking this same graph as \(y=f'(x)\), sketch a possible \(y=f(x)\).

   • f' is positive on (0, π), so f rises to a maximum at x = π
   • f' is negative on (π, 2π), so f falls again
   • f(x) = −2cos x + c, a cosine wave turned upside down

3. (c) Explain why your answer to part (b) is not the only one.

   • f(x) + c has the same gradient as f(x) at every x
   • So the graph can be moved up or down
   • One extra fact, such as a point on the curve, fixes c

Let \(f(x)=x^2(x^2-6)\).

1. (a) Use technology to sketch \(y=f(x)\).

   • From the calculator: graph \(y=x^2(x^2-6)\)
   • Roots −2.45, 0 (touching), 2.45; minima at x = ±1.73

2. (b) Find the values of x for which \(f(x)>0\).

   • Read where the curve is above the x-axis
   • x < −2.45 or x > 2.45

3. (c) Find the values of x for which \(f'(x)<0\).

   • Turning points at x = −1.73, 0, 1.73
   • f falls for x < −1.73 and for 0 < x < 1.73
   • f'(x) < 0 on those intervals

The graph shows four curves, P, Q, R and S. They form two pairs: in each pair, one curve is the derivative of the other. Write down the pairs and say which curve is the derivative.

• P is a quartic with turning points at x = −1, 0 and 1, so its derivative is a cubic with those roots: Q
• R is a downward parabola with maximum at x = 0, so its derivative is 0 at x = 0, positive for x < 0 and negative for x > 0: the line S
• The derivative of P is Q, and the derivative of R is S