The vertical profile of a sculpted roller coaster rail is modeled by the function h(x)=(x−4)(2x+5)2h(x) = (x - 4)(2x + 5)^2h(x)=(x−4)(2x+5)2 for x≥−3x \ge -3x≥−3, where hhh is the height in decimetres and xxx is the horizontal distance from a sensor.
The rail touches the baseline at point PPP and crosses the baseline at point QQQ.
State the coordinates of the point PPP.
Determine h′(x)h'(x)h′(x).
Hence show that the equation of the tangent to the rail at the point where x=116x = \frac{11}{6}x=611 can be expressed in the form y=ky = ky=k, where kkk is a constant to be found.
A modification shifts the track horizontally so the equation becomes y=h(x+b)y = h(x + b)y=h(x+b), where bbb is a constant. The modified track now passes through the sensor's origin O(0,0)O(0,0)O(0,0).
State the possible values of bbb.
717 exam-style questions on Edexcel A Level Maths Differentiation, covering 9.1 Differentiating sin x and cos x, 9.2 Differentiating exponentials and logarithms, 9.3 The Chain Rule, 9.4 The Product Rule, 9.5 The Quotient Rule, 9.6 Differentiating Trigonometric Functions, 9.7 Parametric Differentiation, 9.8 Implicit Differentiation, 9.9 Using Second Derivatives, and 9.10 Rates of Change. Each one has a worked solution and a mark scheme showing where the marks go.