MHB Is the Inflection Point at x = 7 Incorrect?

[riri]

Hello again :)

I am doing a question in which I have to find the mistake that is in one of these points and must change one of these criteria to make it correct. So first I am drawing a graph to get an outline of what it looks like.

-f(x) is defined on (−∞, ∞) and differentiable twice through this domain
-f(x) has critical numbers only at x =−1 and x =3
-f(−5)=0, f(0)=−2, f(1)=0
-f(x) has its only inflection point at x = 7.

I am predicting that the last point, : that the inflection point at x=7 is the wrong one? because when I sketched out the points, I plotted the critical points, and the f(-5)=0, etc, but I don't know how to draw the "twice times differntiable" part...How do I know where I should draw the graph concave up, down?

Thank you! :D

 

[anemone]

Hello again :)

I am doing a question in which I have to find the mistake that is in one of these points and must change one of these criteria to make it correct. So first I am drawing a graph to get an outline of what it looks like.

-f(x) is defined on (−∞, ∞) and differentiable twice through this domain
-f(x) has critical numbers only at x =−1 and x =3
-f(−5)=0, f(0)=−2, f(1)=0
-f(x) has its only inflection point at x = 7.

I am predicting that the last point, : that the inflection point at x=7 is the wrong one? because when I sketched out the points, I plotted the critical points, and the f(-5)=0, etc, but I don't know how to draw the "twice times differntiable" part...How do I know where I should draw the graph concave up, down?

Thank you! :D

Your prediction is correct, but you have to convince your teacher why the inflexion point doesn't fall at $x=7$.

In fact, based on what we've been told, you could first let $f(x)=a(x+5)(x-1)(x+b)$, then you can figure out what the $a$ and $b$ values are, and from there you could determine that the inflexion point is at $x=\dfrac{1}{3}$.

 

[riri]

Hello!

Thank you for your help!
Oh I see! After I drew the graph, I understood that the inflection point is near x=1.
What I'm confused on is how I can find the values of a and b using that equation you wrote?
I tried to expand it, but it clearly got more weird and long so if you could help me with this that would be great

Thank you!

 

[anemone]

We're given that $f(x)$ has critical numbers only at $x =−1$ and $x =3$, that means $f'(-1)=0$ and $f'(3)=0$, see if you could use that to figure out the values for both $a$ and $b$, okay? :D

Edit:

I didn't realized until now that the values for both $a$ and $b$ (that I have found above) wouldn't give us back $f(0)=-2$.

Upon re-reading the given information I have to say this problem is not that well-structured...

Thanks to greg1313 for letting me know there is something wrong about my solution. :D

 

[riri]

Hello! Thank you :)

So basically that inflection point at x=1/3 you proposed can't be true is what you're saying, right?
After I looked at the problem and thought about it more I guess for this problem, it's not possible to give a direct inflection point, just what it should be between...I think?

 

[Greg]

1. f(x) is defined on (−∞, ∞) and differentiable twice through this domain
2. f(x) has critical numbers only at x =−1 and x =3
3. f(−5)=0, f(0)=−2, f(1)=0
4. f(x) has its only inflection point at x = 7.

Assume f(x) is cubic. Assume points 1, 2 and 3 are true and 4 is false (which it must be if f(x) is a cubic as any inflection point must lie between the critical numbers).

Now we need to justify our assumption that f(x) is cubic under conditions 2 and 3:

$$a(x+5)(x-1)(x-b)=0$$
$$a(x^2+4x-5)(x-b)=0$$
$$a(x^3+4x^2-5x-bx^2-4bx+5b)=0$$
$$ax^3+a(4-b)x^2-a(5+4b)x+5ab=0$$
$$5ab=-2\implies ab=-\dfrac25$$
$$f'(x)=3ax^2+2a(4-b)x-a(5+4b)$$
$$f'(-1)=3a-8a+2ab-5a-4ab=0\quad(*)$$
$$f'(3)=27a+24a-6ab-5a-4ab=0\quad(**)$$
The two equations $(*)$ and $(**)$ have no simultaneous solution other than $a=0$, so $f(x)$ cannot be a cubic under conditions 2 and 3.

Have you typed the problem correctly?