# Linear equation is equal to a cubic equations, question about number of solutions?

$f(x) = -x^3 +3x^2 +9x - 11$
$f(x) = 9x + b$
$-x^3 +3x^2 +9x - 11 = 9x + b$

For what values of b does this have three different solutions...
From the looks of it the linear equation needs to cross the cubic one i three places. The rise of the linear equation is 9, is that random?

From the part task before this one I found two equations that I was suppose to use in the solving part of this equation. I just don't see how they help!

$Y = 9x -11$
$Y = 9x - 7$

These are just the equations for the two tangents on f(x) with rise equal to 9.

• Can you use calculus? May 27 '12 at 19:51
• One could expect more from someone with 67k+ in rep. May 27 '12 at 20:27
• The calculus approach happens to be easier here. There is also a more messy algebraic approach. I wanted to know what the prescribed tools are, in order to give a suitable answer. May 27 '12 at 20:36
• Haha. I feel bad now. I read that as insult. That probably says a lot about me. Yes I can use calculus to solve it. But nothing to fancy. I forgot to add another part, two secs. May 27 '12 at 20:40

I interpret the problem as asking the following. Let $f(x)=-x^3+3x^2+9x-11$ and let $g(x)=9x+b$. For what values of $b$ does the equation $f(x)=g(x)$ have three different solutions? Equivalently, for what values of $b$ does the curve $y=f(x)$ meet the line $y=g(x)$ at three different points?

We arrive at the equation $-x^3+3x^2+9x-11=9x+b$. This simplifies to $$x^3-3x^2=-(11+b).$$

We now draw the graph of $y=x^3-3x^2$. Note that $\frac{dy}{dx}=3x^2-6x$. This is $0$ at $x=0$ and $x=2$.

So $x^3-3x^2$ is steadily increasing until $x=0$, where it reaches $0$. Then it steadily decreases until $x=2$, where it reaches the value $-4$. After that, the curve steadily increases.

By looking at the graph, we can see that the horizontal line $y=-(11+b)$ meets the curve $y=x^3-3x^2$ at $3$ distinct places precisely if $-(11+b)$ lies between $0$ and $-4$ (but not including these points). So the $b$'s that work are all $b$ in the open interval $(-11, -7)$.

There are in a sense three points of intersection when $b=-11$, but two happen to coincide. The same is true when $b=-7$. For values of $b$ not in $[-11,-7]$, there is only one point of intersection.

Remark: An equivalent solution that is not as evident geometrically is that we will have three meeting points between the values of $b$ for which the line $y=9x+b$ is tangent to the curve $y=-x^3+3x^2+9x-11$. For tangency, slopes must match, so we want $$-3x^2+6x+9=9,$$ from which we get $x=0$ and $x=2$. In order for $y=9x+b$ to be tangent to $y=f(x)$ at $x=0$, values must match, meaning that $-11=b$. For tangency at $x=2$, values must match, giving $-8+12+18-11=18+b$, meaning that $b=-7$.

• Thanks a lot. I'll be reading over this and drawing the graph myself so I understand it properly. Sorry for being a little bitch at first and thanks for replying anyway. May 27 '12 at 21:46