# Thread: Find the interval in which the function concave up??

1. ## Find the interval in which the function concave up??

The graph of the function

is concave up on the interval:

(-infinity,infinity)

None

(1,infinity)

(-infinity,0)U(1,infinity)

(0,infinity)

=======================

I solve the question and my answer is :

The graph is concave up on (-infinity,-0.5)U(0,infinity)

and I'm sure from my answer but my answer is not included in the choices

and When I use the graphics calculator the graph is concave up on all interval

So what is the correct answer??

2. You can use derivatives to answer this question

First you must find the critical points, take your first equation and differentiate and solve for 0

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

$f\prime(x)=4x^3+3x^3+1$

Now that its differentiated solve for 0

$4x^3+3x^2+1=0$

$x(4x^2+3x+1)=0$

you a critical point at x = -1.5 and x = 0

now to find a graph is increasing you plug your critical points back in the derivative and if its <0 graph is decreasing >0 increasing

to find concavity you take second derivative and plug in critical points if

>0 then concave up and <0 concave down

the second derivative is

$12x^2+6$

plus in 0 and -1.5

plug 0 in and you see it is >0 so intervals from 0 to infinity is concave up

plug -1.5 it is also concave up because > 0 so from -inf. to -1.5

sorry for lack of work studying for finals

3. Originally Posted by sk8erboyla2004
You can use derivatives to answer this question

First you must find the critical points, take your first equation and differentiate and solve for 0

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

$f\prime(x)=4x^3+3x^3+1$

Now that its differentiated solve for 0

$4x^3+3x^2+1=0$

$x(4x^2+3x+1)=0$

you a critical point at x = -1.5 and x = 0

now to find a graph is increasing you plug your critical points back in the derivative and if its <0 graph is decreasing >0 increasing

to find concavity you take second derivative and plug in critical points if

>0 then concave up and <0 concave down

the second derivative is

$12x^2+6$

plus in 0 and -1.5

plug 0 in and you see it is >0 so intervals from 0 to infinity is concave up

plug -1.5 it is also concave up because > 0 so from -inf. to -1.5

sorry for lack of work studying for finals

but I would to remind you that in concavity test we do the following :

1)find first derivative

2)find second derivative

3)set first derivative=0 and solve for x

4)usinig the zeros of second derivative we will set open interval and select test number within each interval to determine the type of concavity ..

4. Originally Posted by change_for_better
The graph of the function

is concave up on the interval:

(-infinity,infinity)

None

(1,infinity)

(-infinity,0)U(1,infinity)

(0,infinity)

=======================

I solve the question and my answer is :

The graph is concave up on (-infinity,-0.5)U(0,infinity)

and I'm sure from my answer but my answer is not included in the choices

and When I use the graphics calculator the graph is concave up on all interval

So what is the correct answer??
A function is concave upward when its second derivative is positive.
If $y= x^4+ x^3+ x$, then $y"= 12x^2+ 6x= 6x(2x+ 1)$. That is 0 at x= 0 and x= -1/2. If x< -1, x< 0 and 2x+1< 0 so y" is positive. If -1< x< 0, x< 0 and 2x+1> 0 so y" is negative. If x> 0, 2x+1> 0 so y" is positive. y is concave upward on exactly the interval you give, whether is one of the options or not!

5. Originally Posted by change_for_better

but I would to remind you that in concavity test we do the following :

1)find first derivative

2)find second derivative

3)set first derivative=0 and solve for x

4)usinig the zeros of second derivative we will set open interval and select test number within each interval to determine the type of concavity ..

If you know this why you couldnt have used this to answer your own question?

6. Originally Posted by HallsofIvy
A function is concave upward when its second derivative is positive.
If $y= x^4+ x^3+ x$, then $y"= 12x^2+ 6x= 6x(2x+ 1)$. That is 0 at x= 0 and x= -1/2. If x< -1, x< 0 and 2x+1< 0 so y" is positive. If -1< x< 0, x< 0 and 2x+1> 0 so y" is negative. If x> 0, 2x+1> 0 so y" is positive. y is concave upward on exactly the interval you give, whether is one of the options or not!

I would like to ask you ,when I use graphics calculator the shape of the graph of the function is concave up in all interval,butwhen we solve the question the interval (-0.5,0)is concave down ??

7. Originally Posted by sk8erboyla2004
If you know this why you couldnt have used this to answer your own question?

Because ,when I use graphics calculator the shape of the graph of the function is concave up in all interval,butwhen we solve the question the interval (-0.5,0)is concave down ??

8. Hello, change_for_better!

I agree with your intervals . . .

The graph of the function: . $f(x) \;=\;x^4+x^3+x$ is concave up on the interval:

. . $(a)\;(-\infty, \infty) \qquad(b)\text{ None} \qquad (c)\;(1,\infty) \qquad (d)\;(-\infty,0) \;\cup\; (1,\infty) \qquad(e)\;(0,\infty)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I solved and my answer is: . $\left(-\infty,-\tfrac{1}{2}\right) \cup (0,\infty)$

and I'm sure from my answer but my answer is not included in the choices.

and when I use the graphics calculator the graph is concave up on all interval
. .
Um ... not quite!

So what is the correct answer?

$\begin{array}{ccc}f(x) &=&x^4+x^3+x \\ f'(x) &=& 4x^3+3x^2+1 \\ f''(x) &=& 12x^2+6x \end{array}$

$f''(x) > 0 \quad\Rightarrow\quad x < \text{-}\tfrac{1}{2}\:\text{ or }\:x > 0$

If you zoom in on your graph, there is a concave-down portion.
Code:
                    |
|         *
|        *
|      *
*               |   *
- - - - - - - - - o - - - - - - - -
*           o  |
*        o    |
*     *     |
*        |
|

And it's right where you predicted! . . . $\left(\text{-}\tfrac{1}{2},\:0\right)$

9. Originally Posted by Soroban
Hello, change_for_better!

I agree with your intervals . . .

$\begin{array}{ccc}f(x) &=&x^4+x^3+x \\ f'(x) &=& 4x^3+3x^2+1 \\ f''(x) &=& 12x^2+6x \end{array}$

$f''(x) > 0 \quad\Rightarrow\quad x < \text{-}\tfrac{1}{2}\:\text{ or }\:x > 0$

If you zoom in on your graph, there is a concave-down portion.
Code:
                    |
|         *
|        *
|      *
*               |   *
- - - - - - - - - o - - - - - - - -
*           o  |
*        o    |
*     *     |
*        |
|
And it's right where you predicted! . . . $\left(\text{-}\tfrac{1}{2},\:0\right)$

Thank you very very very very much Soroban for great explanation

Now I understand my mistake ..

So from choices I should select none because the correct answer is not included ..