Wrong Inflection point

Question

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:

Can someone explain why? Thank You.

Answer 1

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.

Answer 2

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 $(1, 2)$.

Answer 3

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.

Answer 4

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 $y=-3x+5$.

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.

Answer 5

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.

Answer 6

Note the relationships between the first and second derivatives:

$\hspace{5cm}$

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.

$\hspace{4cm}$

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.