4400 Higher November 2005 Paper 4 Q15

EdexcelHigherOld spec8 marksDifferentiation

15 The diagram shows the graph of \(y = x^3 - 12x + 17\)
\(A\) is the maximum point on the curve.
\(C\) is the minimum point on the curve.
The curve crosses the \(y\) axis at \(B\).

Sketch of the cubic curve y = x cubed − 12x + 17 with maximum point A to the left of the y-axis, B where the curve crosses the y-axis, and minimum point C above the x-axis to the right

For the equation \(y = x^3 - 12x + 17\)

(a) find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), (2)

(b) find the gradient of the curve at \(B\), (2)

Scheme	Marks
\(3x^2 - 12\)	B2
	(2)

Scheme	Marks
\(3 \times 0^2 - 12\)	M1
−12	A1
	(2)

Mark scheme (c)

Scheme	Marks
(his \(3x^2 - 12\)) = 0	M1
\((x - 2)(x + 2) = 0\) or \((3x - 6)(x + 2) = 0\) oe	M1
\(x = 2\) or \(-2\) or \((2, 1)\)	A1
\(A\) is \((-2, 33)\), \(C\) is \((2, 1)\)	A1
	(4)
	(8 marks)

Notes

M1: or \(x^2 = 4\)