1. Answer * all* parts (a) to (c).

Consider the function *y*(*x*) = (*x* + 2)^{3}.

(a) Differentiate this function (without multiplying out the brackets) *(4 marks)*

(b) Multiply out the brackets and differentiate it again, confirming that your answer agrees with that obtained in part (a)
*(4 marks)*

(c) Determine the co-ordinates (*x*,*y*) of the stationary point(s). *(8 marks)*

(d) Do the stationary point(s) correspond to local maxima, minima, or point(s) of inflection?
*(8 Marks)*

(e) Hence sketch this function between *x* = -4 and *x* = +1. *(8 marks)*

(f) What is the area beneath this curve between *x* = 2 and *x* = 4 ? *(8 Marks)*

a) Using Function Of Function Rule: *dy/dx* = 3(*x* + 2)^{2}.(1) = 3(*x* + 2)^{2}

b) Multiplying out gives *y* = *x*^{3} + 6*x*^{2} + 12*x* + 8,

so *dy/dx* = 3*x*^{2} + 12*x* +12 = 3(*x* + 2)^{2} as before.

c) At the t.p., *dy/dx* = 0, so 3(*x* + 2)^{2} = 0, so there's only one solution at *x* = -2.

Feeding *x* = -2 back into the original eqn, we get that the t.p is at: (-2, 0).

d) *d*^{2}*y*/*dx*^{2} = 6(*x* + 2), and at *x* = -2 this has a value of 0. So the t.p. is a __point of inflexion__!

Checking the gradient either side of the t.p.: at *x* = -3, *dy/dx* = +3, so it's increasing.
At *x* = -1, *dy/dx* = +3, so it's increasing too. So the t.p. is a +ve pt of infection.

e) When *x* is very large and +ve, *y* is v. large and +ve. When *x* is large and -ve, *y* is large and -ve. When *x* = -2, *y* = 0, and this is the t.p. When *x* = 0, the intercept *y* = 8.

f) We need to integrate it in the expanded form from part (b):

Area =

= (64+128+96+32) - (4+16+24+16) = (320) - (60) =