# Homework Help: Concavity, critical #, second derivatives, extrema (with pictures)

1. Nov 25, 2008

### asdfsystema

hey guys, hope you can help me with this, heres the picture:

1. to find whether it is increasing or decreasing,

f'(x)<0 is decreasing and f'(x)>0 is increasing.

so first-> i took the derivative and got f'(x)=12x^2+60x-168 and set it to zero to find the critical numbers. I got x= -7 and x= 2. Next I plugged it back into the original to find where it is decreasing on the interval,

f(-7)= 4(-7)^3+30(-7)^2-168(-7)+2 = 1276
f(2)= -182

now my question is what do i put in the blanks (the blanks with the decreasing and increasing interval)

I can fill in one blank and that is local maximum at -7.

2. for A and B, i first took the f'(x) and got sqrt(x^2+4) + (x)(x / sqrt(x^2+4).

Next, I took the second derivative and got f"(x)= 2x^2+x+4 / sqrt(x^2+4).

I do not know what to do next. I keep getting confused about this interval stuff. Do i set it to zero?

The point of inflection is when it changes from concave up to down and vice versa, so i find the critical numbers first? how do i get the point of inflection?

For maximum and minimum, i need to find the critical numbers and plug it into the second derivative to see if they are greater, less, or equal to zero?

I'll stop here and do the last problem tomorrow because i feel this is already too much.

Hope I made it as clear as possible. Thanks for all the help !

2. Nov 25, 2008

### HallsofIvy

The value at those points does not tell you whether f is increasing or decreasing on the intervals. Look at the value of f' at a point in each interval.

[/quote]now my question is what do i put in the blanks (the blanks with the decreasing and increasing interval)

I can fill in one blank and that is local maximum at -7.

2. for A and B, i first took the f'(x) and got sqrt(x^2+4) + (x)(x / sqrt(x^2+4).

Next, I took the second derivative and got f"(x)= 2x^2+x+4 / sqrt(x^2+4).

I do not know what to do next. I keep getting confused about this interval stuff. Do i set it to zero?

The point of inflection is when it changes from concave up to down and vice versa, so i find the critical numbers first? how do i get the point of inflection?

For maximum and minimum, i need to find the critical numbers and plug it into the second derivative to see if they are greater, less, or equal to zero?

I'll stop here and do the last problem tomorrow because i feel this is already too much.

Hope I made it as clear as possible. Thanks for all the help ![/QUOTE]

3. Nov 26, 2008

### asdfsystema

Sorry I can't really see what you responded to, can you make it a little bit more clear? thanks

4. Nov 26, 2008

### Staff: Mentor

For problem 4:
You said:
You're mixing up ideas about the function f and its derivative f'.
f(x) is decreasing when f'(x) < 0, and f(x) is increasing when f'(x) > 0.

For this function, you calculated the derivative correctly and solved the equation f'(x) = 0 to get the critical numbers.
What you also needed to do was find out where f'(x) < 0 and where f'(x) > 0.
f'(x) > 0 when 12(x + 7)(x - 2) > 0, which is equivalent to x > 2 or x < -7.
Similarly, f'(x) < 0 when -7 < x < 2.

These two statements tell us that the graph is rising on (neg. infinity, -7), falling on (-7, 2), and rising again on (2, infinity). From that you can see that there must be a local maximum at (-7, f(-7)) and a local minimum at (2, f(2)).

You can use the second derivative to confirm the conclusion above. f''(x) = 24x + 60 = 12(2x + 5).
f''(x) < 0 (graph is concave down) when x < -5/2. f''(x) > 0 (graph is concave up) when x > - 5/2. This is in agreement with the previous work.

Since concavity is changing on either side of x = -5/2 and f''(x) = 0 for x = -5/2, there is an inflection point at (-5/2, f(-5/2)).

For problem 5:
I can't spend much time right now, as I need to get ready for work, but I'll give you some of the high points.

f'(x) = (2x^2 + 4)/(sqrt(x^2 + 4)) This is the best form for doing further work. To find out the intervals on which f is increasing and decreasing, find the intervals for which f'(x) > 0 and f'(x) < 0, respectively. This amounts to solving the inequalities 2x^ + 4 > 0 and 2x^2 + 4 < 0. Luckily for you, 2x^2 + 4 > 0 for all x, which says that the graph of f is increasing over its entire domain, and in particular, on [-5, 4]. This should give you an idea of where the minimum and maximum values have to be.

I calculated f''(x) = (2x(x^2 + 6))/(x^2 + 4)^(3/2).
You can get an idea of the concavity of the graph of f by determining where f''(x) > 0 (graph of f is concave up) and where f''(x) < 0 (graph of f is concave down).
This amounts to solving, respectively 2x(x^2 + 6) > 0 and 2x(x^2 + 6) < 0, which isn't too hard to do (the quadratic factor is always positive).

I'll take a look at #6 later today.

5. Nov 26, 2008

### Staff: Mentor

Problem 6 is very easy. The graph is a parabola opening downward. Any maximum or minimum will occur at a point where f'(x) = 0 or at an endpoint of the interval of definition.

6. Nov 27, 2008

### asdfsystema

Happy Thanksgiving :) !

4. Thank you so much for explaining so clearly. I know what my teacher is talking about and I took a look at this before going to school today and understood everything in class. If it wasn't for your explanation, i would still be completely clueless !

5. may i ask what happened to the sqrt(x^2+4) ? because i see that you're telling me to figure out the inequalities 2x^2+4>0 and 2x^2+4<0 which makes sense but I'm not sure why we're not finding out the entire f'(x)= 2x^2+4 / sqrt(x^2+4) . Could it be because we're only interested in the domain?

following your suggestions, i solved that it will give me (2x-2)(x+2) which means my critical numbers are x=1 and x=-2. Next I plugged both of these back in the original equation which gives me (1, 1sqrt(5)) which will be my minimum and (-2, -2sqrt(8)) which will be my maximum? I hope I am right...

I'm still kind of confused how to determine how long it is going to be concave up/down... I tried solving 2x(x^2+6) setting it to 0 and i know that either 2x=0 or x^2+6 has to be equal to 0 or else it won't work out...

6. I took the derivative and got f'(x)= -6x . set it to 0 and noted that x=0. so i plugged the endpoints back into the original equation and got (-3,23) and (1,1). Does that mean that the absolute max value is 1 and absolute min is -3?

thank you for being so patient. i also messaged you your inbox, please check :)

Last edited: Nov 27, 2008
7. Nov 27, 2008

### Staff: Mentor

You want to solve f'(x) = 0, which happens when the numerator is zero. We can disregard the denominator because x^2 + 4 is always positive, so its square root is always defined.
Check your factors: does (2x - 2)(x + 2) multiply to 2x^2 + 4? If your factors are wrong, that affects any subsequent work.
Right. So f''(0) = 0 or f''(x) = 0 for whatever values of x make x^2 + 6 = 0. For this latter part, think about what the graph of y = x^2 + 6 looks like, and in particular, where this graph crosses the x-axis. Those numbers will be the solutions of the equation x^2 + 6 = 0.

Another way to look at it--for what x is x^2 = -6?
It does as long as f(0) lies between f(-3) and f(1). You did check f(0), right? A max or min occurs at places where f'(x) = 0 or at endpoints of an interval.

8. Nov 27, 2008

### asdfsystema

5.

Okay I tried again this time using the quadratic formula instead, to factor it.

This time I got x= sqrt(-32)/4 and x= sqrt(32)/4 .

Next, I plugged x= sqrt(-32)/4 into the original and got the (x,y) coords: [ (-sqrt(32)/4), -3.4635785 ] , which will be my minimum

Then I plugged x= sqrt(32)/4 into the same original equation and got the (x,y) coords: [ (sqrt(32)/4), 3.4635785 ] which will be my maximum.

For the concave up/down part. First we have to take the second derivative which you showed me, and then set it to 0. (2x)(x^2+6)=0 , so the x will make 2x=0 and will make x^2+6=0.

For x^2+6=0 when x= -2.449, it will equal to 0.
For 2x=0 , when x=0, it will equal to 0.
Since I know this, what can I conclude about the intervals where it concaves up and down?

Next, to find the inflection point, I set f”(x)=0. This time I used the calculator and found that only when x=0 will y=0. I think we can conclude that x=0 is a inflection point?

6. I did f(0) and got f(0)=4 …
When I did f’(x)=0 , I get that f’(0)= -6(0) which will give me (0,0).

For the endpoints, it is (-3,23) and (1,1). I don’t think (4, f(0)) lies on f(-3) and f(1) after I typed it in the calculator…

Ah. I’m running out of time. This assignment is due tomorrow, but I’ll try my best to figure it out

9. Nov 28, 2008

### Staff: Mentor

For #5, we got to f'(x) = (2x^2 + 4)/(sqrt(x^2 + 4)).

You were trying to solve the equation 2x^2 + 4 = 0 using the quadratic formula, and I think you got the coefficients a, b, and c in this formula switched around.

Here a = 2, b = 0, and c = 4, so using the quadratic formula, we get x = (0 +/- sqrt(0 -32)/4 = +/-sqrt(-32) / 4
This is not a real number! This means that there are no real solutions to the equation 2x^2 + 4 = 0. This further means that there are no points on the graph with horizontal tangents.

Max or min points have to occur at the endpoints in this problem.

For concavity, find where f''(x) < 0 and where f''(x) > 0.
From earlier work, this amounts to solving 2x(x^2 + 6) = 0 and analyzing how this expression can be negative or positive.

Of the two factors, x^2 + 6 is always >= 0. In fact, it is always >= 6. So the only factor that can influence the sign of f'' is x.

So...
f''(x) < 0 when x < 0, and f''(x) > 0 when x > 0.
Or graph of f is concave down for x < 0 and is concave up for x > 0.

There have been several simple quadratic expressions in this problem: 2x^2 + 4, x^2 + 4, and x^2 + 6. If you look at all of these as functions, they are parabolas that open upward and have no x-intercepts. Their vertices (their low points) are respectively at (0, 4), (0, 4), and (0, 6). The y-values on these parabolas are always positive.

Any real number squared (or raised to the power 4, 6, 8, or any even power) is always >= 0. Any real number squared (or raised to any even power) + a positive number is always >= that positive number.

10. Nov 28, 2008

### asdfsystema

Thanks Mark . I understand that the minimum is -5 and the maximum is 4 because of the endpoints of the interval. and I also know that the inflection point is at x=0 y=0. But for the concavity, are there specific numbers as to where f(x) is concave down/up on the region from ___ to ___ ?

f"(x) < 0 when x<0 it is concave down
f"(x) > 0 when x>0 it is concave up

I finally understood those, but are there exact values as to where this region occurs and how can i find out? Because that is what my professor is trying to ask me to figure out

11. Nov 28, 2008

### Staff: Mentor

A graph won't be concave up or down over an entire interval. For the function we're talking about, the graph is concave down for x < 0; i.e., (-inf, 0). Similar for where it's concave up.