# Inflection point conc up/down

## Homework Statement

Identify the inflection points, local min/max and intervals the graph is rising, falling, conc up/down analytically, support graphically.

y=-x^3-3x^2-4x-2

## The Attempt at a Solution

F'(y)=-3x^2-6x-4
x=-1+/-.5775i

F"(y)=-6x-6
x=-1

I've done this before and I can do it analytically for roots that aren't imaginary...I also need help with the sign graph for the roots...

Mark44
Mentor

## Homework Statement

Identify the inflection points, local min/max and intervals the graph is rising, falling, conc up/down analytically, support graphically.

y=-x^3-3x^2-4x-2

## The Attempt at a Solution

F'(y)=-3x^2-6x-4
x=-1+/-.5775i
There's no need to be concerned about complex or imaginary roots. You're looking for real values of x for which dy/dx = 0. Your work shows there aren't any.

Since there aren't any x-values for which f'(x)= 0, it can't change sign. Since f'(0) = -4, f'(x) must be negative everywhere. What does that imply about the graph of y = -x3 - 3x2 - 4x - 2?
F"(y)=-6x-6
x=-1

I've done this before and I can do it analytically for roots that aren't imaginary...I also need help with the sign graph for the roots...
Since f''(-1) = 0 and f''(x) changes sign at x = -1, what does that say about the concavity of the graph of y = -x3 - 3x2 - 4x - 2?

1) so it's decreasing from -infinity to -infinity?
2) conc up at x=-1

Also how do I do this problem...
rotate region defined by y=(1/4)x^2 and y=2x+1 at x-axis and find volume...
do I start off by finding intercepts?
x=-.4721 x=8.4721 then do the integral for volume
pi*integral of ((2x+1)^2-((1/4)x^2)^2) from -.4721 to 8.4721?

Mark44
Mentor
1) so it's decreasing from -infinity to -infinity?
It's decreasing from -infinity to +infinity.
2) conc up at x=-1
No, that's where the concavity changes. You're supposed to say on what intervals the graph of the original function is concave up or concave down.

I'm not sure what to say, it's not clear like what we've done in my class before..I've only seen clear ones such as x^2/3 functions where there is a clear dip.

It is concave down when the second derivative is negative.
It is concave up when the second derivative is positive.

Let's say $$f(x)=x^{3}$$.
$$f'(x)=3x^{2}$$
$$f''(x)=6x$$

First derivative tells you whether the function is increasing or decreasing.

Second derivative tells you the concavity. If the second derivative is negative, it is concave down. If the second derivative is positive, it is concave up.

When $$f''(x)=0$$ that is called an inflection point.

To figure out whether whether it's concave up or down we must plug in a point. Let's plug in -1.
$$f''(-1)=6(-1)=-6$$
This tells us this is concave down.

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Mark44
Mentor
Also how do I do this problem...
rotate region defined by y=(1/4)x^2 and y=2x+1 at x-axis and find volume...
do I start off by finding intercepts?
x=-.4721 x=8.4721 then do the integral for volume
pi*integral of ((2x+1)^2-((1/4)x^2)^2) from -.4721 to 8.4721?
That would sort of work, but I would not switch to approximations until the very end. If you start off with approximations your accuracy will be affected.

The integral would be
$$\pi \int_{4 - 2\sqrt{5}}^{4 - 2\sqrt{5}} [(2x + 1)^2 - x^4/16]dx$$

Thanks and I can't see your pictures PiRho.

Thanks and I can't see your pictures PiRho.

Sorry latex fail It's fixed now. If you need more examples, feel free to ask.

x>-1 = -
x<-1 = +
at f"(x)= 6x-6
So I'm confused :/

Let's take the last part of your equation for an example.
$$f(x)=-3x^{2}-4x-2$$
$$f'(x)=-6x-4$$
The root of the first derivative is when $$x=-2/3$$.
So let's plug in -1 and 0 since it's to the left and right of the critical point.
$$f'(-1)=-6(-1)-4=2$$
So from $$-\infty$$ to $$-1$$ it is increasing. Now let's plug in 0.
$$f'(0)=-6(0)-4=-4$$
So from $$-1$$ to $$\infty$$ it is decreasing.

Now let's move on to the second derivative.
$$f''(x)=-6$$
It is always negative for all x. So from $$-\infty$$ to $$\infty$$ it is concave down. If there was a root you would do the same process as you did for the first derivative but instead of it telling you whether the function is increasing or decreasing, it is telling you whether the function is concave up or concave down in that portion.

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no, the -6x-6 was the second derivative.

no, the -6x-6 was the second derivative.

Notice my $$f(x)$$ is different from your problem. I'm making up a whole new problem. I was hoping you take take my example and figure out your problem.

So then is it just still concave down?

So then is it just still concave down?

As you mentioned $$f''(x)=-6x-6=-6(x-1)$$. So there is a root at $$x=1$$.

Let's look at when $$f''(0)$$ and when $$f''(2)$$.

$$f''(0)=-6(0)-6=-6$$ which means concave up.
$$f''(2)=-6(2)-6=-12-6=-6$$ which means concave down.

Ohhh ok sorry, but thank you so very much. Still getting use to some problems haha.

Ohhh ok sorry, but thank you so very much. Still getting use to some problems haha.

Yup. No problem. Feel free to ask. I'm struggling with math myself :-)

Mark44
Mentor
As you mentioned $$f''(x)=-6x-6=-6(x-1)$$. So there is a root at $$x=1$$.
No, f''(x) = -6x - 6 = -6(x + 1).
Let's look at when $$f''(0)$$ and when $$f''(2)$$.

$$f''(0)=-6(0)-6=-6$$ which means concave up.
Actually, since f''(0) = -6 < 0, it means the original graph is concave down.
$$f''(2)=-6(2)-6=-12-6=-6$$ which means concave down.

We have f''(x) = -6(x + 1), with f''(-1) = 0

For x < - 1, f''(x) > 0 so the graph of f is concave up on this interval.
For x > - 1, f''(x) < 0 so the graph of f is concave down on this interval.

Since the concavity changes at x = -1, there's an inflection point at (-1, f(-1)).

No, f''(x) = -6x - 6 = -6(x + 1).
Actually, since f''(0) = -6 < 0, it means the original graph is concave down.

We have f''(x) = -6(x + 1), with f''(-1) = 0

For x < - 1, f''(x) > 0 so the graph of f is concave up on this interval.
For x > - 1, f''(x) < 0 so the graph of f is concave down on this interval.

Since the concavity changes at x = -1, there's an inflection point at (-1, f(-1)).

Thanks for correcting me. It was really late last night :-)

Mark44
Mentor
Thanks for correcting me. It was really late last night :-)
It's an easy mistake to make, with all those minus signs...