# Slopes of tangents of a cubic function

Recently I came across this problem where it was stated:

"A cubic function y=ax3+bx2+cx+1 has only one tangent which has the slope of 4, and which touches the graph in x=-1/3"

I did the best I could to translate it into English. The problem went on ofcourse, but this is the part which raised the following question in my head:
shouldnt a cubic function like this have 2 tangents with the same slope in every 2 according points on the graph, EXCEPT for the extremum point?
Where do I go wrong?

When you solved the problem, you got that y=3x3+9x2+9x+1
You can easily draw it out on http://rechneronline.de/function-graphs/

## Answers and Replies

tiny-tim
Homework Helper
hi fawk3s!
"A cubic function y=ax3+bx2+cx+1 has only one tangent which has the slope of 4, and which touches the graph in x=-1/3"

When you solved the problem, you got that y=3x3+9x2+9x+1
but that doesn't go through x = 1/3

I think you missed the minus there

tiny-tim
Homework Helper
I think you missed the minus there
ooh, yes

that doesn't go through x = -1/3

Now Im confused. What doesnt go through -1/3?

tiny-tim
Homework Helper
y=3x3+9x2+9x+1

If x=-1/3, then y=3*(-1/3)3+9*(-1/3)2+9*(-1/3)+1=-10/9
So the tangent is in the point (-1/3; -10/9)

It does, why shouldnt it?

tiny-tim
Homework Helper
sorry, i've misread the question

yes, x = -1/3 would have to be an inflexion point, and have a slope of 4 …

how did you get that result?​

I did mention above that I didnt write out all of the problem. So you are missing some info here. The problem goes:

"There's a cubic function y=ax3+bx2+cx+1, which has only one tangent with a slope of 4, and that tangent touches the given function in x=-1/3. It is also known that the cubic function has extremum in x=-1. Find a,b and c."

Im not really familiar with all the terms in English so I dont know if my translation makes sense.

Anyways, I solved it by constructing a "function system" (dont know how its called in English ), which contains
y'(-1/3)=4
y'(-1)=0

And I had trouble getting that last third function to complete that "function system", so I used
y''(-1)=0

because I assumed that the funcion has only one extremum point and that its neither maximum nor minimum point. This actually confuses me, because say you take a function like
y=3x3+10x2+9x+1

and you get 2 extremums, of which one is maximum, and the other minimum.

I think I got lucky, because the given answer is a=3, b=9, c=9. But Im rather confused if I solved it correctly, because I made that assumption.

But now to my real question:
doesnt a function like y=3x3+9x2+9x+1 have 2 tangents with identical slopes, except for the one in the extremum (assuming there's only one extremum)?

Im rather confused here acutally.

tiny-tim
Homework Helper
… And I had trouble getting that last third function to complete that "function system", so I used
y''(-1)=0

I think I got lucky, because the given answer is a=3, b=9, c=9.
shouldn't it be y''(-1/3)=0 ?

(and i get a = b = -c = 6 )

No, by my logic, the extremum, which is in x=-1, is neither a minimum nor a maximum, and therefore
y''(-1)=0

Im not sure if this is correct though since it involves an assumption that the given extremum is the only extremum the function has. I'll ask my teacher tomorrow for confirmation.
As I can see you have taken interest in the problem like I have, so I'll be sure to post what my teacher teacher will say tomorrow.

But what do you think about my original question, the one I created this thread for?

fawk3s

tiny-tim
Homework Helper
No, by my logic, the extremum, which is in x=-1, is neither a minimum nor a maximum, and therefore
y''(-1)=0
i don't follow this

at an extremum, y' is 0, not y''

i don't follow this

at an extremum, y' is 0, not y''
Maybe my logic does fail here. I'll try and explain:

Say we have an extremum in x. So
y'(x)=0

If y''(x)>0, then the extremum is a minimum.
If y''(x)<0, then the extremum is a maximum.

Say for a function like y=x3, there is only one point where y'(x)=0. But this point is neither a minimum or maxiumum (take a look at the graph). So I figured that if
y''(x) cannot be bigger or smaller than 0, then it must be 0.
Therefore y''(x)=0

Is this a fail in my logic?

It did give me the right answer, so I figured it must be true, but Im not really satisfied with my solution, because it does involve an assumption about the extremum. I think Im missing something easier and more logical here which I could use as the third function.

tiny-tim
Homework Helper
if y''(x) cannot be bigger or smaller than 0, then it must be 0.
your argument is faultless

but i don't see where your premise comes from

I indeed have to admin I was wrong. My version holds true for the function I brought out, but you really cant make an assumption like that based on nothing - if it solves true, its just blind luck. Plus, my function did not hold true for the given fact that the function has only one tangent with the slope of 4. So I failed to solve it correctly.

It was easy to solve actually, but I missed something fairly simple. Since the problem states that the function has one and only one tangent with a slope of 4, then y'=4 has only one solution, which means the discriminant D=0.
So when y'=3ax2+2bx+c, then
3ax2+2bx+c=4
3ax2+2bx+c-4=0

Solving for x,
x=-2b+-sqrt[4b2-4*3a*(c-4)]/2*3a
from which D=0 is
4b2-4*3a*(c-4)=0

which would be the last function needed.

I also missed the obvious fact that you were indeed correct.
y''(-1/3)=0 also holds beautifully true.

HallsofIvy