# Finding extremas and points of inflection

• Nov 25th 2012, 07:27 PM
Chaim
Finding extremas and points of inflection
Find all relative extrema and points of infelction
Answers are: Relative minimum: (1, -27), Points of infelction: (2, -16), (4, 0)

f(x)=x(x-4)3
It could also be simplified into x4-64x
f'(x)=3x(x-4) OR 4x3-64
0=3x(x-4)2 + 3x OR 0=x4-64x
Thus, x = 0, and x = 1
f''(x)=6x(x-4) + 3x + 6x OR 12x2
f''(0)=6(0)(0-4) + 3(0) + 6(0)=0 so here is not a relative extrema
f''(3)=6(3)(3-4) + 3(3) + 6(3)=9 (I think I think the product rule wrong for the original equation, and second derivative, if someone could help me fix that, so I'll use the other function instead to get it)
f''(3)=12(1)2 = 1 Here is the relative minima
So I plugged it back into the original equation and got the relative extrema to be (1, -27)

Now finding the Points of Inflection has been my problem
I went around and saw that people say to have the second derivative set equal to 0
0=12x2
0=x
Then they say to plug that point into the original equation.
f(x) = 0(0-4)3 = 0
And that was not one of the points of inflection in the answer.

Can anyone explain? Thanks :)
• Nov 25th 2012, 07:38 PM
MarkFL
Re: Finding extremas and points of inflection
Your first step contains an error, you have:

\$\displaystyle (x-4)^3=x^3-4^3\$

and this is not true. This type of error is so commonly made that it is sometimes referred to as "the freshman's dream."

I would suggest that instead of using the expansion, you use the product rule to differentiate the original function instead.
• Nov 25th 2012, 08:00 PM
Chaim
Re: Finding extremas and points of inflection
Quote:

Originally Posted by MarkFL2
Your first step contains an error, you have:

\$\displaystyle (x-4)^3=x^3-4^3\$

and this is not true. This type of error is so commonly made that it is sometimes referred to as "the freshman's dream."

I would suggest that instead of using the expansion, you use the product rule to differentiate the original function instead.

Um ok,
Can you see if my product rule is correct
f(x) = x(x-4)3
f'(x) = (3(x-4)2)(x) + (x-4)3(1) = 3(x-4)2(x)+ (x-4) = Something that I think I did wrong o.o
f''(x) = Something I haven't quite figured out yet due to the first derivative.

The product rule is like (derivative of inside)(outside) + (inside)(derivative of outside) right?
I kind of got lost due to the exponent outside.
• Nov 25th 2012, 08:15 PM
MarkFL
Re: Finding extremas and points of inflection
You have correctly differentiated in your first step, but in the next step you dropped the exponent of 3 on the second term. You should have:

\$\displaystyle f'(x)=3x(x-4)^2+(x-4)^3\$

Now, since you will be equating this to zero, you want to factor it fully, so that you may apply the zero-factor property to find the roots, which will be the critical number with respect to extrema.
• Nov 25th 2012, 08:52 PM
Chaim
Re: Finding extremas and points of inflection
Quote:

Originally Posted by MarkFL2
You have correctly differentiated in your first step, but in the next step you dropped the exponent of 3 on the second term. You should have:

\$\displaystyle f'(x)=3x(x-4)^2+(x-4)^3\$

Now, since you will be equating this to zero, you want to factor it fully, so that you may apply the zero-factor property to find the roots, which will be the critical number with respect to extrema.

Ah ok,
So that means it'll be 3x(x2-8x+16) + (x+4)3
Which equals 3x3-24x2+48x + (x+4)3
Which equals to (3x3-24x2+48x) + (x3-12x2+48x-64)
Which equals to 3x2-24x2+48x+x3-12x2+48x-64
Which equals to 4x3-36x2+96x-64
Which equals to 4(x3-9x2+24-16)
So... how do I factor this thing down.
Isn't there an easier way than this instead of FOILing, then FOILing again (FOIL is to distribute everything from the outside into the inside and the other way around), and then simplifying, then simplifying it down again. This seems like a lot of work for a single problem, and I might easily mess up on one thing that could affect the whole.

So, is there an easier way than this one? :) And also, if I did do this correctly, how would I keep factoring it down to get the relative extremas?

Then that means the second derivative is:
12x2-72x+96

Thanks.
• Nov 25th 2012, 08:55 PM
MarkFL
Re: Finding extremas and points of inflection
To factor the first derivative, I would begin by noticing that both terms have \$\displaystyle (x-4)^2\$ as a factor...
• Nov 25th 2012, 09:12 PM
Chaim
Re: Finding extremas and points of inflection
Quote:

Originally Posted by MarkFL2
To factor the first derivative, I would begin by noticing that both terms have \$\displaystyle (x-4)^2\$ as a factor...

Oh oops, my bad
Would it be like this? (3x+1)(x-4)2+(x-4)?
(3x+1)(x-4)2 -> (3x+1)(x2-8x+16)+(x-4)
3x3-24x2+48x+x2-8x+16+x-4
• Nov 25th 2012, 09:23 PM
MarkFL
Re: Finding extremas and points of inflection
No, I would write:

\$\displaystyle f'(x)=(x-4)^2(3x+(x-4))=(x-4)^2(4x-4)=4(x-4)^2(x-1)\$

Now, recall from you pre-calculus studies what a root of even multiplicity implies. Actually, you don't really need to consider this as the second derivative test for the nature of the function at the critical points will show you the same thing, but it is good to be aware of these things going into the test.

So, you are armed with two critical points, and since you are instructed to also find points of inflection, compute the second derivative using the product rule, and you may then use the second derivative both to determine the nature of the possible extrema and to find the points of inflection.
• Nov 26th 2012, 02:37 PM
Chaim
Re: Finding extremas and points of inflection
Oh! I see now.
So the Critical numbers are then 4, and 1 right?
And the second derivative would be 2(x-4)(4x-4)+(x-4)2(4) which equals to (2x-4)(4x-4)+4(x-4)2 which then turns into 12x2-56x+80
Um, did I do this correct?
• Nov 26th 2012, 03:18 PM
MarkFL
Re: Finding extremas and points of inflection
I would compute the second derivative as follows:

\$\displaystyle f''(x)=4((x-4)^2(1)+2(x-4)(x-1))=4((x-4)((x-4)+2(x-1))=\$

\$\displaystyle 4(x-4)(3x-6)=12(x-4)(x-2)\$

Since no roots are repeated, we know the points of inflection are:

\$\displaystyle (2,f(2)),\,(4,f(4))\$

Now, what does the second derivative test tell you about the critical points found from the first derivative?
• Nov 26th 2012, 03:28 PM
Chaim
Re: Finding extremas and points of inflection
Oh oops, forgot about the 4 outside,
Well since we have the second derivative, knowing about the critical numbers that were found from the first one, we see that the critical numbers were 4 and 1.
So that means when we plug the critical numbers into the second derivative, f''(4) would have no relative extrema since it equals to 0 in the second derivative test, while if we plug in f''(1) = 12(-3)(-1) = 36. That would mean this is the relative minimum since it's greater than 0

So then when we plug in the point 1 back into the original equation x(x-4)^3, then it would be 1(1-4)^3, so that means it's (1, -27)

But anyway the points of inflection are: 2, 2(2-4)^3 which is (2, -16), while for 4 would be (4, 0) since the 4 makes everything else a 0
Ah! So I see! So the points of inflection are like 'critical numbers' of the second derivative? Whatever makes it a 0?

Now I see it!!! Thanks! :)