Question

The graph of f(x) = – x³ + 27x – 2 is given below:

1. Find the slope of the above graph.   [1]
2. Find the co-ordinates of turning points, A and B.   [2]
3. Evaluate f"'(–2), f(0) and f'(3) and arrange them in ascending order.   [1]

Answer 1

f(x) = –x³ + 27x – 2   ...(i)

a. Slope = f'(x) 

Slope f'(x) = –3x² + 27   ...(ii)

b. f'(x) = 0

–3x² + 27 = 0

x² = 9

⇒ x = ± 3

Substitute x = ± 3 in equation (i), we get at x = 3

f(3) = y

= – (3)³ + 27 × 3 – 2

= – 27 + 81 – 2

= 52

At x = –3

f(–3) = y 

= –(–3)³ + 27 × (–3) – 2

= + 27 – 81 – 2

= –56

Therefore, the coordinates of turning points of A and B are A(–3, –56) and B(3, 52).

c. Again, differentiate w.r.t. x equation (ii)

f"(x) = –6x

Now, f"(–2) = –6 × (–2)

= 12

f'(3) = –3(3)² + 27

= –27 + 27

= 0

f(0) = –(0)³ + 27 × 0 – 2

= –2

f"(–2) = 12, f'(3) = 0 and f(0) = –2

Arranging in ascending order –2, 0, 12.