0
$\begingroup$

I have the next function: $f(x)=x^3-3x^2+4$.
I need to find an inflection point so I had done the following steps:

I found the first derivative: $f'(x)=3x^2-6x$

Then the second one: $f''(x)=6x-6$

Comparative to zero: $6x-6=0$
$6x=6\rightarrow x=1$

I checked values before and after the point in the second derivative:
$f''(0)=-6<0$
$f''(2)=12>0$
So the point (1,2) is inflection point but when I chaked the graph, there were not any:
enter image description here


Can someone explain why? Thank You.

$\endgroup$
  • $\begingroup$ The first derivative gives you critical points not the second derivative. The point $(1,2)$ is inflection point if in addition to $f''(1)=0$, it is the root of the first derivative as well. $\endgroup$ – Yadati Kiran Nov 24 '18 at 10:56
  • $\begingroup$ Yes, I know. But I need an inflection point and not critical points. $\endgroup$ – violettagold Nov 24 '18 at 10:59
  • $\begingroup$ So there are none in this case. $\endgroup$ – Yadati Kiran Nov 24 '18 at 11:00
  • $\begingroup$ What do you mean by root? $\endgroup$ – violettagold Nov 24 '18 at 11:02
  • 2
    $\begingroup$ 1,2 is inflection point $\endgroup$ – Akash Roy Nov 24 '18 at 11:26
2
$\begingroup$

Draw the graph together with the tangent at $(1,f(1))$: $1$ is where the second derivative vanishes and $f(1)=2$; since $f'(1)=-3$, the tangent has equation $$ y-2=-3(x-1) $$ or $y=-3x+5$.

enter image description here

As you see, the tangent “crosses” the graph, because the curve is concave for $x<1$ and convex for $x>1$.

So at $x=1$ there's indeed an inflection point.

$\endgroup$
  • $\begingroup$ Thank you very much for the help! $\endgroup$ – violettagold Nov 24 '18 at 12:03
1
$\begingroup$

In order to find out critical points first find the points where derivative equals to zero and check the concavity of graph about that point.

For example if $0$ is an inflection point,

Then $f^{''}(0^{+})>0$ and $f^{''}(0^{-})<0$ or vice - versa.

$\endgroup$
  • $\begingroup$ But this is what I got no? So why there is no point on the graph? $\endgroup$ – violettagold Nov 24 '18 at 11:08
  • $\begingroup$ You have got this but your conclusion is wrong $\endgroup$ – Akash Roy Nov 24 '18 at 11:19
  • 1
    $\begingroup$ Hey the point is an inflection point , view it closely by zooming it out, you can check it then. (1,2) is an inflection point $\endgroup$ – Akash Roy Nov 24 '18 at 11:22
  • $\begingroup$ Why is my answer downvoted? $\endgroup$ – Akash Roy Nov 24 '18 at 11:25
  • $\begingroup$ I thought so too but look at the graph, it shows there is no inflection point. $\endgroup$ – violettagold Nov 24 '18 at 11:30
1
$\begingroup$

You’re right, the graph changes its concavity at at exactly the point $x$ where $$f’’(x) = 0$$

such that $f’’(x_1) > 0$ results in concave upward and $f’’(x_1) < 0$ results in concave downward.

$$f’’(x) = 0 \implies 6x-6 = 0 \implies x = 1$$

Which gives $y = 2$, hence the point is $(1, 2)$.

If you notice, the tangent’s slope begins to increase (become less negative and eventually positive at $x > 2$) after the inflection point.

If it's not easy to vizualize it that way, take a look at https://www.desmos.com/calculator/4zu7w0kmd2 and see how the behavior of the tangent line and the graph's concavity both change at point $(1, 2)$.

$\endgroup$
  • $\begingroup$ You are confusing the notion of an inflection point and a local extremum (maximum or minimum) $\endgroup$ – ip6 Nov 24 '18 at 11:26
  • $\begingroup$ Sorry, I misunderstood the question for a second, I’ve edited it. (I didn’t read through the part which confused the OP.) $\endgroup$ – KM101 Nov 24 '18 at 11:34
  • $\begingroup$ I’ve removed the downvote $\endgroup$ – ip6 Nov 24 '18 at 11:35
  • $\begingroup$ Thank you very much for the help! $\endgroup$ – violettagold Nov 24 '18 at 12:04
1
$\begingroup$

An inflection point is where the 2nd derivative changes sign. This is where the curvature changes from concave downward to concave upward or vice versa. It’s where the tangent changes sides.

You have correctly calculated the position of the inflection point - note the inflection point should not be confused with a maximum or minimum.

Draw some tangents on your graph and see what happens around the point you identified.

$\endgroup$
  • $\begingroup$ I am sorry for bothering but can you give me an example, please? $\endgroup$ – violettagold Nov 24 '18 at 11:31
  • 2
    $\begingroup$ Cup your left hand around the shape of the left hand side of the curve - your hand will be palm down. Cup your right hand around the shape of the right hand part of the curve - your hand will be palm up. In the middle, where your fingers meet is a point where the curvature switches. This is the inflection point. And you found it correctly!! $\endgroup$ – ip6 Nov 24 '18 at 11:42
  • $\begingroup$ Now @violettagold you should upvote the answers , $\endgroup$ – Akash Roy Nov 24 '18 at 12:01
  • $\begingroup$ Thank you very much for the help! $\endgroup$ – violettagold Nov 24 '18 at 12:03
  • $\begingroup$ Your first statement is wrong, consider $f(x)=x^4$. $\endgroup$ – Michael Hoppe Nov 24 '18 at 12:37
0
$\begingroup$

I zoomed the central part (for you to see it more clearly).

Inflection point PI is at the crossed reticles $(1,2)$ red lines intersection.

To the right of PI water holds and at left water spills in the cubic curve and so your checked second derivative signs are OK.

The software does not mark the point automatically that is all. The way to recognize an inflection point graphically is that the curve is locally straight, you are neither turning to your right nor to your left as happening elsewhere. This happens when you are asked to drive along an $S$ shaped figure of $8$ curve at center that is your PI.

Cubic PI

$\endgroup$
0
$\begingroup$

Note the relationships between the first and second derivatives:

$\hspace{5cm}$enter image description here

1) $f'<0, f''>0$ - the function is decreasing at an increasing rate;

2) $f'>0, f''<0$ - the function is increasing at a decreasing rate;

3) $f'<0, f''<0$ - the function is decreasing at a decreasing rate;

4) $f'>0, f''>0$ - the function is increasing at an increasing rate.

Now note that the function $f(x)=x^3-3x^2+4$ is decreasing at an increasing rate in $(0,1)$ and decreasing at a decreasing rate in $(1,2)$. And at the point $x=1$, the function changes its rate of decrease.

$\hspace{4cm}$enter image description here

Be careful, $f''(x)=0$ does not imply the point of inflection (e.g. $f(x)=x^4$), so it is a possible inflection point. The function must keep decreasing or increasing around the point of inflection. In the above link you can also see the animated graph of tangent line.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.