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In the above question, I'm stuck on figuring out how to apply the implicit function theorem to solve it.

In part(b), I rewrote $$g(x,y)=x(\varepsilon)^2+2y(\varepsilon)^2+2x(\varepsilon)y(\varepsilon)-2x(\varepsilon)-2y(\varepsilon)+\varepsilon x(\varepsilon)^5$$

Then I figured I would differentiate $g$ with respect to Epsilon and hopefully find an equation where $x'(\varepsilon)=y'(\varepsilon)=0$ when $x(\varepsilon)=1$ and $y(\varepsilon)=0$, satisfying the requirements for $(1,0)$ to be a critical point. However, when differentiating, I ended up with $1+5\varepsilon x'(\varepsilon)=0$, which was very close! (but not quite...)

My other idea was to do the partial derivative with respect to $x$ and another with respect to $y$ and find $x'(\varepsilon)$ and $y'(\varepsilon)$ to be zero, but I don't think I can do that as $x$ and $y$ are now functions (of $\varepsilon$) and not actually variables.

I'm pretty stuck here and sure that I'm differentiating correctly

For part (c), I'm not really sure what to do.

For part (d) and (e), I'm sure I should calculate the Hessian somehow and show that $(x(\varepsilon),y(\varepsilon))$ corresponds to a min, but I'm not sure how to make that differentiation between the two.

I feel like I have a good grasp on this stuff but I'm stuck, so any help would be great. Thanks!



We can calculate that

$$ \frac{\partial g_\varepsilon}{\partial x} = 2x + 2y -2 + 5 \varepsilon x^4, $$ $$ \frac{\partial g_\varepsilon}{\partial y} = 4y + 2x -2. $$

When $\varepsilon = 0$, the two partial derivatives given above are zero at $(x,y)=(1,0)$. The partial derivatives are continuously differentiable and have a non-singular Jacobian at $\varepsilon = 0$. Hence, by the implicit function theorem, $g_\varepsilon$ has a critical point $(x(\varepsilon),y(\varepsilon))$ near $(1,0)$.


$(x(\varepsilon),y(\varepsilon))$ is defined implicitly by

$$2x + 2y -2 + 5 \varepsilon x^4 = 0, $$ $$4y + 2x -2 = 0. $$

We can differentiate the above two expressions by $\varepsilon$ to find

$$ 2x^\prime + 2y^\prime +5x^4 + 20 \varepsilon x^3x^\prime = 0, $$ $$ 4y^\prime +2x^\prime = 0 .$$

Here, $x^\prime$ and $y^\prime$ are the derivatives of $x$ and $y$ with respect to $\varepsilon$. We can then solve these two expressions for $x^\prime$ and $y^\prime$ to find

$$ y^\prime = -\frac{1}{2}x^\prime, $$ $$ x^\prime = -\frac{5x^4}{1 + 20 \varepsilon x^3}. $$


$$ \left.\frac{\mathrm dx(\varepsilon)}{\mathrm d \varepsilon}\right\rvert_{\varepsilon = 0} = -5, $$ $$ \left.\frac{\mathrm dy(\varepsilon)}{\mathrm d \varepsilon}\right\rvert_{\varepsilon = 0} = \frac{5}{2}. $$


The Hessian matrix of $g_\varepsilon$ is given by

$$ \left[\begin{matrix} 2 + 20 \varepsilon x^3 & 2 \\ 2 & 4 \end{matrix}\right].$$

This matrix is positive definite at $(1,0)$ for $\varepsilon = 0$, and thus must be positive definite at $(x(\varepsilon),y(\varepsilon))$ for all $\varepsilon$ close enough to zero. Therefore $(x(\varepsilon),y(\varepsilon))$ must be a local minimum for all $\varepsilon$ sufficiently small.


No. For $\varepsilon \neq 0$, $g_\varepsilon$ is not bounded below.

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  • $\begingroup$ Makes perfect sense!!! $\endgroup$ – Anthony Mar 26 '18 at 21:22

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