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anonymous
 one year ago
The figure at below is the graph of f ′′ (x), the second derivative of a function f (x). The domain of the function f (x) is all real numbers, and the graph shows
f ′′ (x) for −2.6≤ x ≤3.6.
anonymous
 one year ago
The figure at below is the graph of f ′′ (x), the second derivative of a function f (x). The domain of the function f (x) is all real numbers, and the graph shows f ′′ (x) for −2.6≤ x ≤3.6.

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A. Find all values of x in the interval (−2.6, 3.6) where f ′(x) has a horizontal tangent. B. Find all values of x in the interval (−2.6, 3.6) where f (x) is concave upwards. Explain your answer. C. Suppose it is known that in the interval (−3.6, 3.6), f (x) has critical points at x =1.37, and x = −0 .97. Classify these points as relative maxima or minima of f (x). Explain your answer.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2what do you have so far?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Don't know where to start.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2A. Find all values of x in the interval (−2.6, 3.6) where f ′(x) has a horizontal tangent.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2f ' (x) has a horizontal tangent whenever the tangent slope on f ' (x) is 0 that only happens when f '' (x) = 0

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2It's very similar to saying "f(x) has a horizontal tangent at the point where f ' (x) = 0"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So would occur at the maximums and minimums?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2no I'm saying that the horizontal tangent on f ' (x) happens at the roots of f '' (x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So how do we find those?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2what are the roots of f '' (x)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you're given the graph

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the answer would be 2, 1 , 3?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2for part A, yes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That was easier than I thought.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2at those x values, the value of f '' is 0, so that's where the horizontal tangent on f ' will be

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now how do we do part B?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2f(x) is concave up whenever f '' (x) > 0

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2again you're given the graph of f '' so just list the interval(s) when f '' is above the x axis to report the interval(s) when f is concave up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well I don't think I can from the graph the exact values of f''

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2again they want the interval along the x axis

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2look at the graph. when is f '' (x) > 0 ? what x values?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2list it as an interval

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2from x = ??? to x = ???, the value of f '' (x) is positive

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2good, where else?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well its either that or x=3 to x = we don't know

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2but it gives you the bounds on which f '' is restricted

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2"x = 2 to x = 1" can be written in interval notation as (2,1) (2,1) is NOT a point, it's an interval. yeah the notation is confusing sometimes

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.23 to 3.6 can be written as (3,3.6)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2put the two together with a union symbol and you get \[\Large (2,1) \cup (3,3.6)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2that represents where f '' (x) > 0, where f is concave up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait, isn't concave up when the curve looks like a U not an upside down U? So doesn't that mean it is concave up between x=1 and x=3 only? Is that what the above expression means?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you are correct, but you're thinking of f having those qualities (of having a U or upside down U)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you're given the graph of f '' and they want to know info about f

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2we don't know what f looks like but we can use f '' to figure out when f is concave up or down

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I see. I forgot the graph shows f''(x).

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2That's a common trick calc teachers use, so watch out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, know how do we do part C?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2the critical point x = 1.37 where is this point located on f '' ? is it in a concave up region? or concave down?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2or put another way, if x = 1.37, is f '' positive or negative?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2since f '' is negative when x = 1.37, this means that x = 1.37 lies in a concave down region on f

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2so this is what the small piece of f looks like dw:1433375975175:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2swapped them lol

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2dw:1433376035069:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2dw:1433376059680:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2relative max, yeah

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and 0.97 is the minima for being positive

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually it stays the same, it is the relative minima

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2if x = 0.97, then f '' (x) is positive making f concave up here so if x = 0.97, then there is a relative min on f at x = 0.97 you are correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, thank you for your help. Do you mind helping me with another problem if I do @jim_thompson5910 on the question?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2sure, one more
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