# Thread:

1. ## Concavity / Points of Inflection

I am stuck on this problem, and I am thinking that I am not getting the derivatives (1st and 2nd) correct. Here is the problem:

F(X)= 2X(X+4)^3

I know I have to get the first and second deriviative's but don't I need to cube the inside function first and then distribute the 2x to it? Or does the power rule come into effect here? I have tried cubbing the inside terms and then distributing the 2x but it doesn't seem to give me the correct zeros's of the graph so if someone could illustrate. Please Help, Thanks.

2. Originally Posted by kdogg121
I am stuck on this problem, and I am thinking that I am not getting the derivatives (1st and 2nd) correct. Here is the problem:

F(X)= 2X(X+4)^3

I know I have to get the first and second deriviative's but don't I need to cube the inside function first and then distribute the 2x to it? Or does the power rule come into effect here? I have tried cubbing the inside terms and then distributing the 2x but it doesn't seem to give me the correct zeros's of the graph so if someone could illustrate. Please Help, Thanks.
Let's go through the derivatives together...

$f(x)=2x(x+4)^3$

Apply the product rule

$f'(x)=2(x+4)^3+6x(x+4)^2$

Let's do it again!

$f''(x)=6(x+4)^2 + 6(x+4)^2+12x(x+4) \Rightarrow 12(x+4)[(x+4)+x] = (12x+48)(2x+4)$

3. I'm sorry for the first derivative how are you getting the 6X? Is this the power rule being used within the product rule?

How would I determine where the function is increasing and decreasing, where its concave up and down and the inflection points. I know you have to know the zero's of the derivative(s).

4. Originally Posted by kdogg121
I'm sorry for the first derivative how are you getting the 6X? Is this the power rule being used within the product rule?
The derivative of (x + 4)^3 is 3(x + 4)^2.

Personally, since you need to solve f''(x) = 0 (and, of course, test the resulting solutions), I'd expand f(x) first and then differentiate (twice) term-by-term. I think that's the easier road to hoe.

5. So are you saying it would be easier to use the product rule? I still don't see where the 6X(X+4)^2 comes from, but how would I determine my zero's from here which will result in the x points on the graph?

My book says the function is increasing for x> -1 decreasing for x< -1 concave upward for x<-4 and x> -2 concave downward at -4< x < -2 minimum at (-1) and inflection at (-4,0) and (-2,-32)

Could someone show how I find these?

6. Originally Posted by kdogg121
So are you saying it would be easier to use the product rule? I still don't see where the 6X(X+4)^2 comes from, but how would I determine my zero's from here which will result in the x points on the graph?
I've just re-read my reply and it's pretty plain what I said: "Personally ... I'd expand f(x) first ... I think that's the easier road to hoe."

I cannot see how you could possibly think I said it would be easier to use the product rule.

On the topic of the product rule, do you understand how it works in this question? In particular, do you understand that the second term is 2x times 3(x + 4)^2? Is it becoming clearer now where the 6X(X+4)^2 comes from?

Edit: Your potential x-coordinates for the stationary points will come form solving a quadratic equation. I assume solving a quadratic is money for jam for you ....

7. Originally Posted by kdogg121
[snip]
My book says the function is increasing for x> -1 decreasing for x< -1

Mr F says: Increasing when f'(x) > 0. Decreasing when f'(x) < 0. Your derivative function (which is a cubic) factorises into two simple linear factors (one of which is repeated) ......

concave upward for x<-4 and x> -2 concave downward at -4< x < -2

Mr F says: Concave up when f''(x) > 0. Concave down when f''(x) < 0. In each case you'll be solving a quadratic inequality. Note: The quadratic has two simple linear factors.

minimum at (-1)

Mr F says: Stationary points are found by solving f'(x) = 0. Note: The derivative (which is a cubic) factorises into two simple linear factors (one of which is repeated). Nature is found using either the sign test or the double derivative test.

and inflection at (-4,0) and (-2,-32)

Mr F says: Potential inflection points are found by solving f''(x) = 0. You'll be solving a quadratic equation. Note: The quadratic has two simple linear factors. The nature of these solutions must then be tested. Inflection points correspond to turning points of the derivative .....
Note: One of the inflection points is a stationary point of inflection. The other one is non-stationary.

Could someone show how I find these? Mr F says: Done. Please post if there are details you still can't manage.
..
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