Question

please help me with the concavity test....
so the question states: use the concavity test to determine the intervals on which the graph of the function is concave up and concave down
y=4x^3+21x^2+36x-20

Answer 1

Hey there, to know the concavity of the function we need to calculate the second derivative of that function. Did you do that ?

Answer 2

yeah

Answer 3

its 24x+42

Answer 4

where x=1.75 but then i don't know what to do

Answer 5

Yeah, that's your inflection point. A function f is concave up on an open interval if f'' is positive on the interval, and it's concave down on an open interval if f'' is negative on that interval.

Answer 6

You already did the first step by solving 24x + 42 = 0

Answer 7

yeah and then i got x=1.75 and y=128.75

Answer 8

By the way it's x=-1.75, and then you plug it in the initial equation to get y. That should give you the inflection point. So according to the concavity test, f is concave up when x>-1.75, and it's concave down when x<-1.75

Answer 9

Tell me what you get for y

Answer 10

@arrdasha still here ?

Answer 11

sorry... why is x negative?

Answer 12

wait never mind... i get it

Answer 13

so its -40.125

Answer 14

Yeah ! Good job

Answer 15

now what?

Answer 16

when i do my number line to find the concavity do i put it into the original function, f' or f''

Answer 17

Original function. And that's it you got the answer, f is concave up on ]-1.75 ; +infty[, and it's concave down on ]-infty ; -1.75[

Answer 18

how do you know that??

Answer 19

That's because the concavity and second derivative test states that a function is concave up on an open interval if its second derivative is positive on that interval, and it's concave down on an open interval if its second derivative is negative on that interval.

See : http://www.millersville.edu/~bikenaga/calculus/conc/conc.html

Answer 20

thank

Answer 21

Anytime :)