Please could some one check this and help where required:

The Function:

$\displaystyle f(x) = x^3 + 6x^2 -15x -24$

a) Find its stationary points

So, $\displaystyle f ' (x) = 3x^2 + 6(2x) - 15(1) = 3x^2 + 12x - 15 = 3(x^2 + 4x - 5) = 0$

The equation $\displaystyle f ' (x) = 0 $ is equivilent to: $\displaystyle 3(x^2 + 4x - 5)$ that is $\displaystyle 3(x + 5)(x - 1)$

So the Stationary points are $\displaystyle X = -5$ and $\displaystyle X = 1$

b) Classify the LH Stationary point with 1st Derivative Test

To classify we take X = -5 and choose the test point XL = -6 then we have:

$\displaystyle f ' (XL) = f ' (-6) = 3(-6 + 5)(-6 - 1) = 21 > 0$

c) Classify the RH Stationary point with 2nd Derivative Test

So,$\displaystyle f '' (x) = 3(2x) + 12 = 6x + 12 = 6(x + 2)$

The stationary point XR = 1, so the test point will be XR = 2

$\displaystyle

f '' (2) = 6(2 + 2)= 24 > 0$

So f has a local minimum at X = 1

d) Find the coordinates of each stationary point

so,

$\displaystyle XL - f (-5) = -125 + 150 - (-75) - 24 = 76$

$\displaystyle XR - f (1) = 1 + 6 - 15 - 24 = -32$

Hence the graph passes through the point (-5,76) & (1,-32)

Thanks for your help

A