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.

- NathanJHW

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- NathanJHW

##### 1 Attachment

- NathanJHW

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.

- NathanJHW

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## More answers

- jim_thompson5910

what do you have so far?

- NathanJHW

Don't know where to start.

- jim_thompson5910

A. Find all values of x in the interval (−2.6, 3.6) where f ′(x) has a horizontal tangent.

- jim_thompson5910

f ' (x) has a horizontal tangent whenever the tangent slope on f ' (x) is 0
that only happens when f '' (x) = 0

- jim_thompson5910

It's very similar to saying "f(x) has a horizontal tangent at the point where f ' (x) = 0"

- NathanJHW

So would occur at the maximums and minimums?

- jim_thompson5910

no I'm saying that the horizontal tangent on f ' (x) happens at the roots of f '' (x)

- NathanJHW

So how do we find those?

- jim_thompson5910

what are the roots of f '' (x)

- jim_thompson5910

you're given the graph

- NathanJHW

Oh duh. -2,1,3

- jim_thompson5910

yep

- NathanJHW

So the answer would be -2, 1 , 3?

- NathanJHW

- jim_thompson5910

for part A, yes

- NathanJHW

That was easier than I thought.

- jim_thompson5910

at those x values, the value of f '' is 0, so that's where the horizontal tangent on f ' will be

- NathanJHW

Now how do we do part B?

- jim_thompson5910

f(x) is concave up whenever f '' (x) > 0

- jim_thompson5910

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

- NathanJHW

Well I don't think I can from the graph the exact values of f''

- NathanJHW

- jim_thompson5910

again they want the interval along the x axis

- jim_thompson5910

look at the graph. when is f '' (x) > 0 ? what x values?

- NathanJHW

-1 and 0

- jim_thompson5910

list it as an interval

- jim_thompson5910

from x = ??? to x = ???, the value of f '' (x) is positive

- NathanJHW

x=-2 to x=1?

- jim_thompson5910

good, where else?

- NathanJHW

x=1 to x=3

- jim_thompson5910

you sure?

- NathanJHW

Well its either that or x=3 to x = we don't know

- jim_thompson5910

but it gives you the bounds on which f '' is restricted

- NathanJHW

oh so x=3 to x=3.6

- jim_thompson5910

yep

- jim_thompson5910

"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

3 to 3.6 can be written as (3,3.6)

- jim_thompson5910

put the two together with a union symbol and you get
\[\Large (-2,1) \cup (3,3.6)\]

- jim_thompson5910

that represents where f '' (x) > 0, where f is concave up

- NathanJHW

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?

- jim_thompson5910

you are correct, but you're thinking of f having those qualities (of having a U or upside down U)

- NathanJHW

I'm confused.

- jim_thompson5910

you're given the graph of f '' and they want to know info about f

- jim_thompson5910

we don't know what f looks like
but we can use f '' to figure out when f is concave up or down

- NathanJHW

Oh I see. I forgot the graph shows f''(x).

- jim_thompson5910

That's a common trick calc teachers use, so watch out

- NathanJHW

Alright, know how do we do part C?

- jim_thompson5910

the critical point x = 1.37
where is this point located on f '' ? is it in a concave up region? or concave down?

- NathanJHW

concave up

- jim_thompson5910

or put another way, if x = 1.37, is f '' positive or negative?

- NathanJHW

negative

- jim_thompson5910

since f '' is negative when x = 1.37, this means that x = 1.37 lies in a concave down region on f

- NathanJHW

so it's a maxima?

- jim_thompson5910

so this is what the small piece of f looks like
|dw:1433375975175:dw|

- jim_thompson5910

oops my bad

- jim_thompson5910

swapped them lol

- jim_thompson5910

|dw:1433376035069:dw|

- jim_thompson5910

|dw:1433376059680:dw|

- NathanJHW

so it's the maxima?

- jim_thompson5910

relative max, yeah

- NathanJHW

and 0.97 is the minima for being positive

- NathanJHW

oops -0.97

- NathanJHW

actually it stays the same, it is the relative minima

- jim_thompson5910

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

- NathanJHW

Alright, thank you for your help. Do you mind helping me with another problem if I do @jim_thompson5910 on the question?

- jim_thompson5910

sure, one more

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