How should I show that there exists only one solution?

[Math100]

Homework Statement

Consider the equation $g(a)=\Gamma$ where $g(a)=(c-1)a+\frac{3}{4}a^3$.
Show that for $c\geq 1$, only one solution exists for $\Gamma>0$.

Relevant Equations

None.

So far, I've got that $g(a)=(c-1)a+\frac{3}{4}a^3\implies g'(a)=c-1+\frac{9}{4}a^2$.
I know that if the first derivative of a function is positive (greater than $0$),
then that function is always/strictly increasing. However, how should I construct
this proof in order to show that there exists only one solution with the given quantifiers of
$\Gamma>0, c\geq 1$? What needs to be done?

 

[person_random_normal]

When a tends to negative infinity, the function tends to negative infinity, and when a tends to positive infinity, the function tends to positive infinity. Therefore, since the function's first derivative is always positive, the given equation will have only one solution.

 

[Math100]

Thank you for the help!

 

[vela]

You need to be a bit more careful. If $c=1$, which is allowed, then $g'(0) = 0$, so the function isn't always strictly increasing. You should deal with this case as well in your proof.

 

[WWGD]

You need to be a bit more careful. If $c=1$, which is allowed, then $g'(0) = 0$, so the function isn't always strictly increasing. You should deal with this case as well in your proof.

Problem statement specifies $c \geq 1$

 

[vela]

Problem statement specifies $c \geq 1$

What's your point?

 

[WWGD]

If $c=1$, then $f'(a)=(9/4)a^2$, strictly increasing, or constant if $a==0$. Edit: Never mind, you're right. The case $a==0$ should be dealt with. My bad.

 

[pasmith]

I suspect you mean *real* solutions.

The discriminant of the depressed cubic x^3 + px + q is \Delta = -4p^3 - 27q^2.
If p and q are real, then the cubic has three distinct real roots if \Delta > 0 and a real root and a complex conjugate pair if \Delta < 0. If \Delta = 0 then the cubic has a repeated root.

For \frac34 a^3 + (c-1)a = \Gamma we have <br /> \Delta = -\left( \frac 43 \right)^2 \left( \frac{16}{3}(c-1)^3 + 27\Gamma^2\right). For c \geq 1 and \Gamma &gt; 0, we can see that \Delta &lt; 0 so there is a single real root.

 

[mathwonk]

The statement in post #4 that since g'(0) = 0 when c=1, then g is not strictly increasing, is false. In this case g(a) is a constant multiple of a^3, which is strictly increasing everywhere. The definition of strictly increasing is not that the derivative is positive, but rather that g(s) > g(t) whenever s>t. I.e. it is true that a function with everywhere positive derivative is strictly increasing, but the converse statement is not true, and g(a) = a^3 is the standard counterexample.
This is a very easy error to make, even one of my math professor friends initially made it in a calculus textbook he wrote.

I believe one can check that, as in our situation, a non - constant polynomial f with derivative f'≥0 everywhere, is everywhere strictly increasing. The point is that then the derivative can only be zero a finite number of times. (A weakly increasing function that is not strictly increasing, would take the same value twice, and then must be constant on the interval joining those two points.)

 

[SammyS]

The statement in post #4 that since g'(0) = 0 when c=1, then g is not strictly increasing, is false. In this case g(a) is a constant multiple of a^3, which is strictly increasing everywhere. The definition of strictly increasing is not that the derivative is positive, but rather that g(s) > g(t) whenever s>t. I.e. it is true that a function with everywhere positive derivative is strictly increasing, but the converse statement is not true, and g(a) = a^3 is the standard counterexample.
This is a very easy error to make, even one of my math professor friends initially made it in a calculus textbook he wrote.

@vela didn't exactly say that in Post #4. To quote that post:

You need to be a bit more careful. If $c=1$, which is allowed, then $g'(0) = 0$, so the function isn't always strictly increasing. You should deal with this case as well in your proof.

(My emphasis)

In other words, since $g'$ does take on a value of zero, OP needs to take into account further considerations before stating that $g$ is strictly increasing.

 

[mathwonk]

Forgive me if I was unclear. The claim is that in fact any differentiable function g with g'(x) > 0 for all x ≠ 0, is always strictly increasing. No hypothesis on the value g'(0) is required. The point is that the usual mean value theorem argument gives strict increasing (for a differentiable hence continuous function) on the closed interval [a,b] even if the derivative is positive only on the open interval (a,b). Hence it gives here that g is strictly increasing both for x≤0 and for x≥0, hence everywhere.

The definitive statement of the criterion for strict increasing occurs on page 204 of the book Calculus of one variable, by Joseph Kitchen: A differentiable function f on an interval with f'≥0 everywhere is weakly increasing, and is strictly increasing unless f'=0 throughout some open (sub)interval. (In particular, vanishing of f' at isolated points is irrelevant.)

 

[vela]

Forgive me if I was unclear.

I don't think you were unclear. My issue was with the claim in post #2 that $g'>0$ everywhere even when $c=1$. My error was saying that $g$ therefore wasn't strictly increasing. As you noted, $g$ satisfies the definition of strictly increasing even when $c = 1$.

The OP just needed to do a little more work in his proof to establish that $g$ is strictly increasing for the case $c=1$. I suspect that's why the condition $\Gamma > 0$ was included in the problem.