Question

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.

Answer 1

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.

Answer 2

@jim_thompson5910

Answer 3

what do you have so far?

Answer 4

Don't know where to start.

Answer 5

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

Answer 6

f ' (x) has a horizontal tangent whenever the tangent slope on f ' (x) is 0

that only happens when f '' (x) = 0

Answer 7

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

Answer 8

So would occur at the maximums and minimums?

Answer 9

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

Answer 10

So how do we find those?

Answer 11

what are the roots of f '' (x)

Answer 12

you're given the graph

Answer 13

Oh duh. -2,1,3

Answer 14

yep

Answer 15

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

Answer 16

@jim_thompson5910

Answer 17

for part A, yes

Answer 18

That was easier than I thought.

Answer 19

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

Answer 20

Now how do we do part B?

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

@jim_thompson5910

Answer 25

again they want the interval along the x axis

Answer 26

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

Answer 27

-1 and 0

Answer 28

list it as an interval

Answer 29

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

Answer 30

x=-2 to x=1?

Answer 31

good, where else?

Answer 32

x=1 to x=3

Answer 33

you sure?

Answer 34

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

Answer 35

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

Answer 36

oh so x=3 to x=3.6

Answer 37

yep

Answer 38

"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

Answer 39

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

Answer 40

put the two together with a union symbol and you get

\[\Large (-2,1) \cup (3,3.6)\]

Answer 41

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

Answer 42

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?

Answer 43

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

Answer 44

I'm confused.

Answer 45

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

Answer 46

we don't know what f looks like

but we can use f '' to figure out when f is concave up or down

Answer 47

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

Answer 48

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

Answer 49

Alright, know how do we do part C?

Answer 50

the critical point x = 1.37

where is this point located on f '' ? is it in a concave up region? or concave down?

Answer 51

concave up

Answer 52

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

Answer 53

negative

Answer 54

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

Answer 55

so it's a maxima?

Answer 56

so this is what the small piece of f looks like