Question

For what intervals is the function f(x) = x^3 – 2x^2 + 3x – 5 concave down?
A. The graph is concave down for the interval (-∞, 0).
B. The graph is concave down for the interval (-∞,∞ ).
C. The graph is concave down for the interval (-∞,2/3 ).
D. The graph is concave down for the interval (2/3, -∞ ).

Answer 1

Did you find the first and second derivatives?

Answer 2

yes first derivative is 3x^2 - 4x + 3 and second derivative is 6x - 4

Answer 3

Do the same thing as before. Find the points where the first derivative is 0 and put them on your number line.

Answer 4

i got 0 and 1

Answer 5

Did you solve 3x^2-4x+3=0 ?

Answer 6

yes

Answer 7

Well. replacing x with 0 does not make it true and replacing x with 1 does not make it true. In fact. There are no real zeros.

Answer 8

ok

Answer 9

So having discovered that, go to the second derivative and set it equal to 0 and solve.

Answer 10

okay

Answer 11

well i get x = 2/3

Answer 12

So that divides the number line into two parts. Choose an number from each side of 2/3 to see if it is positive or negative.

Answer 13

Is that what you got?

Answer 14

yes

Answer 15

So we see that the function is concave downward to the left of 2/3 and concave upward to the right of 2/3 and we write:
The function is concave downward from (-infinity to 2/3) and concave upward from (2/3, infinity)

Answer 16

And then we check the graph to make sure we are correct:
http://www.wolframalpha.com/input/?i=f%28x%29+%3D+x^3+%E2%80%93+2x^2+%2B+3x+%E2%80%93+5