1. Of the function

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)

A

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 $\displaystyle 3(x + 5)(x - 1)$
You mean $\displaystyle 3(x^2+4x- 5)= 0$ and $\displaystyle 3(x+5)(x-1)= 0$, of course.

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:

$\displaystyle f ' (XL) = f ' (-6) = 3(-6 + 5)(-6 - 1) = 21 > 0$
Just checking at one point is not in general sufficient. Here it is- if you knew that, you should have said how you knew. Also, you haven't answered the question! You haven't said what kind of point that is.

f'= 3(x-1)(x+5). If x is close to -5, then x-1 is close to -5-1= -4 and so is negative. For x< -5, x+ 5< 0 so f'= 3(x-1)(x+5) is the product of one positive and two negative numbers- f' is positive so f is increasing coming up to -5. For x> -5, x+5> 0 so f'= 3(x-1)(x+5) is the product of two positive and one negative number. f'< 0 so f is decreasing past -5. Since f was increasing and then started decreasing, x= -5 is a maximum.

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
$\displaystyle f '' (2) = 6(2 + 2)= 24 > 0$
No, no, no! For the second derivative test, you evaluate f" at the stationary point! f"(1)= 6(1+2)= 3> 0.

So f has a local minimum at X = 1
Yes, but you were lucky!

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)

A
Why are you taking points exactly "1" away from the stationary point? There is nothing special about integers!

3. Originally Posted by HallsofIvy

No, no, no! For the second derivative test, you evaluate f" at the stationary point! f"(1)= 6(1+2)= 3> 0.

Yes, but you were lucky!

Why are you taking points exactly "1" away from the stationary point? There is nothing special about integers!
So for the Second Der. test, how do you make it 3? I make it 18?

4. Originally Posted by HallsofIvy
Why are you taking points exactly "1" away from the stationary point? There is nothing special about integers!
Well to work out the y coordinates of each the stationary point dont we use -5 and 1 ?