Show that implicit function is negative.

Question

Let $f(x,y)=(1+3x^2-\sin x)y+3^{3y}=0$ this function defines a unique implicit function $\phi:\mathbb{R}\to\mathbb{R}$ show that $\phi<0$.

I plotted the graph of this implicit function on Grapher ($\phi<0$ indeed), but what I still don't understand is how to study the behavior of the function. Any help?

I tried to draw the graph by hand but this seems hard to me.

Answer 1

Note that $f(x,0)=1$ for any $x.$ So on $f(x,y)=0$ all $y\neq 0.$ So we get that:

$$1+3x^2-\sin x = -\frac{3^{3y}}{y}$$

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Claim: $1+3x^2-\sin x> 0$ for all real $x$

Proof: Since $\sin x\leq 1$ so $$1+3x^2-\sin x=3x^2+(1-\sin x)\geq 3x^2$$ When $x=0$ then $1+3x^2-\sin x = 1>0,$ and when $x\neq 0,$ $3x^2>0.$

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This means that $y$ must always be negative, because the sign of $3^{3y}$ is always positive.

Now, if $g(y)=\frac{3^{3y}}{y}$ then $g$ is one-to-one on the negative numbers. This is because:

$$g'(y)=g(y)\left(\log 27 - \frac{1}{y}\right)$$

When $y$ is negative, $g(y)$ is negative and $log 27 - \frac{1}{y}>0$ so $g'(y)<0.$

Thus, $g$ is $1-1$ and strictly decreasing on the negative numbers. You can show:

$$\begin{align}\lim_{y\to -\infty}g(y)&=0\\ \lim_{y\to 0^-} g(y) &= -\infty.\end{align}$$ So $g$ is one-to-one and onto from the negative reals to the negative reals.

So we define a function $$\phi(x)=g^{-1}(\sin x-1-3x^2).$$

Then $y=\phi(x)$ if and only if $f(x,y)=0.$

I don't think there is a "nice" closed form for $g^{-1},$ nor for $\phi.$

Answer 2

To begin with, let us rearrange the given equation as it follows \begin{align*} (1+3x^{2} - \sin(x))y + 3^{3y} = 0 \Longleftrightarrow y = -\frac{3^{3y}}{1+3x^{2}-\sin(x)} \end{align*}

Since both the numerator and the denominator are strictly positive, we conclude $y(x) < 0$ due to the negative sign.

Answer 3

A rough outline: we can start by rearranging terms to try and get something that's closer to an 'explicit' function: $3^{3y} = -y(1+3x^2-\sin x)$, or $3^{3y}/y = -(1+3x^2-\sin x)$. Now, since $3^{3y}$ is always positive, note that the LHS of this equation has the same sign as $y$ itself; so the problem comes down to showing that $1+3x^2-\sin x$ is always positive. Can you see how you might show this?