Question

Determine the open intervals on which the graph of the function is concave upward or concave downward. (Enter your answer using interval notation. If an answer does not exist, enter DNE.)
f(x) = −5squarootx

Answer 1

\[f(x)=-\sqrt{x}\] right ?

Answer 2

no, -5 squareroot of x

Answer 3

ok but that won't change the answer
you can do this without any calc if you know what it looks like
you know what \(y=\sqrt{x}\) looks like ?

Answer 4

umm.. not really

Answer 5

i just graphed it, so i do now

Answer 6

oh then you can't do it that way
but it would surely help
graph it with wolfram or a calculator
i would use wolfram
but actually i am fairly sure you do know what it looks like

Answer 7

if you graph \(y=\sqrt{x}\) you see that is it defined for \([0,\infty)\) and increasing on the entire domain
for \(y=-\sqrt{x}\) rotate around the \(x\) axis
domain doesn't change, but this time it is decreasing on the entire domain

Answer 8

i think i have been at this calculus too long today so my mind is kinda fried at this point.. :/ I am giving this up after this problem and coming back to it tomorrow.. anyways.. would the interval notation look something like, ( or[ -inf, 0 ) or ]?

Answer 9

so for open intervals on concave down it is: [0, -inf) and for concave up it is: [0, inf) ?

Answer 10

decreasing on \((0,\infty)\) never increasing

Answer 11

and it is concave up "leaning left, holding water, smiling" on the entire domain as well

Answer 12

you must be tired
there is no such interval is \([0,-\infty)\)

Answer 13

yeah, i am.. i have been working on calculus for 3 days now straight and physics in between with all types of equations.. my mind is so fried... :/ so up: (0,inf) down (0, inf) ??

Answer 14

also if you know power rule, you can do this algorythmically using it. -5sqrt[x]=-5x^1/2

Answer 15

\[-5\sqrt{x}=-5x ^{1/2}\]
\[D[x^{n}]=n*x^{n-1}\]

Answer 16

the derivative is not necessary for this problem
it is concave up and decreasing on its domain
but if you want the derivative then since the derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\) them tje derivative of \(=5\sqrt{x}\) is \(\frac{-5}{2\sqrt{x}}\)

Answer 17

the derivative is not necessary only because you know if by heart, but if you had no idea what it looked like (like zydic) you should definitely use it as a fool proof method of finding concavity.

Answer 18

so if i found derivative i could use that to help see the concavity? it helps to see things visually for me because this stuff is like a foriegn language to me :p

Answer 19

you should do what is best for you, however if you need to do it without a calculator on an exam, take the derivative, then take the derivative of the derivative (second derivative).

If that is greater than 0 the function is concave up on the interval if it is less than zero it is concave down on the interval. its a bit confusing now, but later on this will become invaluable if you dont know what a function looks like

Answer 20

i can always use my calculator on exams, they are online.. so whatever tools i have i use! :) but i am very visual in my learning.. it's just hard for me to understand all this since it is online and i have no formal instruction on how to solve these things.. Anyways, you have been helping me on here.

Answer 21

haha well say you can't tell just from looking at a graph, then you'd need to use this analytic way of doing it. ofc satellite did an excellent job describing it and I completely support learning visually if thats best but sometimes you can't tell from looking

Answer 22

right so i got (0,inf) and (0,inf) .. on concave up and concave down? am i getting that right?

Answer 23

the second derivative is (5/4)x^-3/2

This is always positive when x goes from 0->inf but doesn't include 0
Thus the function is concave up on (0,inf)

Answer 24

I'm done for the night.. thank you sylbot :) thank you sattelite