Question

Use the graph of f "(x) below to state x-coordinates of the inflection points for the graph of f(x).

Answer 1

I got -1 and 5

Answer 2

is that correct?

Answer 3

Mind showing me how you got those two points? I'd suggest we review the concept of "inflection points." The graph illustrated in this problem doesn't have an inflection point, by the way. Can you complete this sentence? "An inflection point is a point on the graph of a polynomial where the 2nd derivative changes ______________." At such a point, the direction of the ____________ of the graph changes (e. g., from up to down or from down to up).

Answer 4

from being concave to convex ?? changes from up to down??

Answer 5

Does that graph ever change from concave to convex? Be careful with "up to down".

Answer 6

observe the curve intersecting x-axis ,
how many points does the curve intersecting can write those points?

Answer 7

It is correct

Answer 8

What is correct lovelarap?

Answer 9

The answer she gave, -1 and 5. It's correct.

Answer 10

curve intersecting at two points (-1 ,0) and (5, 0)

what are the x-coordinates of these points?

Answer 11

OK

Answer 12

Let's recall the concept,

Suppose you have a (differentiable) function in a form of,
\(\color{#000000 }{ \displaystyle y=f(x) }\)

The (possible) inflection points are found the following way,
\(\color{#000000 }{ \displaystyle y''=f''(x) }\)
\(\color{#000000 }{ \displaystyle 0=f''(x) }\)

And suppose that the x-solutions for this is
\(\color{#000000 }{ \displaystyle x=\varphi }\)
(I like greek letters; don't worry, it's just a real number constant)

Then, \(\color{#000000 }{ \displaystyle x=\varphi }\) is a possible inflection point, and the reason
I say "possible" because there is one more testing to verify
whether or not \(\color{#000000 }{ \displaystyle x=\varphi }\) is not indeed an inflection point.

Let's choose a sufficiently small value (I will call it \(\color{#000000 }{ \displaystyle \delta }\)), and
let's verify that the concavity actually changes about \(\color{#000000 }{ x=\varphi }\).

IFF \(\color{#000000 }{ \displaystyle f\left(\varphi+\delta\right) }\) and \(\color{#000000 }{ \displaystyle f\left(\varphi-\delta\right) }\) have different signs (one of them is
positive and the other is negative - doesn't matter which is which),
THEN, concavity (which is determined by the 2nd derivative) indeed
changes about \(\color{#000000 }{ x=\varphi }\), and then \(\color{#000000 }{ x=\varphi }\) is an inflection point.

HOWEVER,
IFF \(\color{#000000 }{ \displaystyle f\left(\varphi+\delta\right) }\) and \(\color{#000000 }{ \displaystyle f\left(\varphi-\delta\right) }\) are both positive or both negative,
THEN, the concavity hasn't changed about \(\color{#000000 }{ x=\varphi }\), and thus \(\color{#000000 }{ x=\varphi }\)
is NOT inflection point.

──────────────────────────────

Now, back to your problem.

You are given the graph of the 1nd derivative \(\color{#000000 }{ f''(x) }\).

I bet you can find the point(s) where \(\color{#000000 }{ f''(x)=0 }\) (the
x-intercepts of the 2nd derivative).

After you see the point(s) that satisfy \(\color{#000000 }{ f''(x)=0 }\), you
can verify the change in concavity, by seeing whether
\(\color{#000000 }{ f''(\varphi \pm \delta) }\) have different signs.

So for example, if my inflection was \(\color{#000000 }{ x=3}\), then, to
verify that \(\color{#000000 }{ x=3}\) is inflection point, you need to find
that \(\color{#000000 }{ f''(3 \pm 0.1) }\) have different signs
(\(\color{#000000 }{ 0.1 }\) is the sufficiently closer \(\delta\) I (might) use)...