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A. 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.
what do you have so far?
Don't know where to start.
A. Find all values of x in the interval (−2.6, 3.6) where f ′(x) has a horizontal tangent.
f ' (x) has a horizontal tangent whenever the tangent slope on f ' (x) is 0 that only happens when f '' (x) = 0
It's very similar to saying "f(x) has a horizontal tangent at the point where f ' (x) = 0"
So would occur at the maximums and minimums?
no I'm saying that the horizontal tangent on f ' (x) happens at the roots of f '' (x)
So how do we find those?
what are the roots of f '' (x)
you're given the graph
Oh duh. -2,1,3
So the answer would be -2, 1 , 3?
for part A, yes
That was easier than I thought.
at those x values, the value of f '' is 0, so that's where the horizontal tangent on f ' will be
Now how do we do part B?
f(x) is concave up whenever f '' (x) > 0
again 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
Well I don't think I can from the graph the exact values of f''
again they want the interval along the x axis
look at the graph. when is f '' (x) > 0 ? what x values?
-1 and 0
list it as an interval
from x = ??? to x = ???, the value of f '' (x) is positive
x=-2 to x=1?
good, where else?
x=1 to x=3
Well its either that or x=3 to x = we don't know
but it gives you the bounds on which f '' is restricted
oh so x=3 to x=3.6
"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
3 to 3.6 can be written as (3,3.6)
put the two together with a union symbol and you get \[\Large (-2,1) \cup (3,3.6)\]
that represents where f '' (x) > 0, where f is concave up
Wait, 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?
you are correct, but you're thinking of f having those qualities (of having a U or upside down U)
you're given the graph of f '' and they want to know info about f
we don't know what f looks like but we can use f '' to figure out when f is concave up or down
Oh I see. I forgot the graph shows f''(x).
That's a common trick calc teachers use, so watch out
Alright, know how do we do part C?
the critical point x = 1.37 where is this point located on f '' ? is it in a concave up region? or concave down?
or put another way, if x = 1.37, is f '' positive or negative?
since f '' is negative when x = 1.37, this means that x = 1.37 lies in a concave down region on f
so it's a maxima?
so this is what the small piece of f looks like |dw:1433375975175:dw|
oops my bad
swapped them lol
so it's the maxima?
relative max, yeah
and 0.97 is the minima for being positive
actually it stays the same, it is the relative minima
if 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
Alright, thank you for your help. Do you mind helping me with another problem if I do @jim_thompson5910 on the question?
sure, one more