1. ## Question about the concavity of graphs

Find where the graph is concave up and concave down.
3X^4 - 16X^3 + 24x^2 + 1

I determined that there were points of inflection at x=2/3 and x=2. At x=2/3 the graph is concave up. At x=2 , the graph is concave down.
If I wanted to state where the graph is concave up or concave down, would it be correct to say that the graph is concave up on the interval (2/3, 2) and concave down on the interval (2, 00]

2. Originally Posted by Kenneth
Find where the graph is concave up and concave down.
3X^4 - 16X^3 + 24x^2 + 1

I determined that there were points of inflection at x=2/3 and x=2. At x=2/3 the graph is concave up. At x=2 , the graph is concave down.
If I wanted to state where the graph is concave up or concave down, would it be correct to say that the graph is concave up on the interval (2/3, 2) and concave down on the interval (2, 00]
Hi Kenneth!

I worked the problem out for you please click on the following 2 links:

1. http://item.slide.com/r/1/92/i/Gpz_9...C2_-tk731Iv0i/
2. http://item.slide.com/r/1/92/i/76ODL...OhHtxGKu0sIch/

Best of Luck!

3. Originally Posted by qbkr21
Hi Kenneth!

I worked the problem out for you please click on the following 2 links:

1. http://item.slide.com/r/1/92/i/Gpz_9...C2_-tk731Iv0i/
2. http://item.slide.com/r/1/92/i/76ODL...OhHtxGKu0sIch/

Best of Luck!
oh, the number line. I understand now.

Thanks.

4. Hello, Kenneth!

Find where the graph is concave up and concave down:
. . f(x) .= .3x^4 - 16x³ + 24x² + 1

I determined that there were points of inflection at x = 2/3 and x = 2 . . . yes!
At x = 2/3, the graph is concave up . . . no
At x = 2 , the graph is concave down . . . no
At a point of inflection, f''(x) = 0.
. . The curve is neither concave up nor concave down.
An inflection point is where the concavity changes.

If I wanted to state where the graph is concave up or concave down,
would it be correct to say that the graph is concave up on the interval (2/3, 2)
and concave down on the interval (2, ∞)> . . . no
The two inflection points create three intervals to test.

. . - - - - * - - - - - * - - - -
. . - - - -2/3- - - - -2

The second derivative is: .f''(x) .= .12(3x² - 8x + 4)

On (-∞, 2/3), try x = 0
. . f''(0) .= .12(3·0² - 8·0 + 4) .= .+48 . concave up: U

On (2/3, 2), try x = 1
. . f''(1) .= .12(3·1² - 8·1 + 4) .= .-12 . concave down: ∩

On (2, ∞), try x = 3
. . f''(3) .= .12(3·3² - 8·3 + 4) .= .+84 . concave up: U

Therefore: .concave up on (-∞, 2/3) U (2, ∞)
. . . . . . . . concave down on (2/3, 2)