Question

Consider the function \(f:\mathbb{R}^2 \to \mathbb{R}\) defined by
\[ f(x_1,x_2)=2x_1^4+x_2^2+x_2x_1^2. \]
Which of the following statements is correct?

• A. There are two distinct stationary points of \(f\)
• B. \((0,0)\) is a saddle point of \(f\)
• C. \((0,0)\) is a local maximum of \(f\)
• D. \((0,0)\) is a local minimum of \(f\)

Hint:

Write \(f\) as a quadratic in \(x_2\) and complete the square in \(x_2\); the result is a sum of two squared terms, so \(f\geq 0\) everywhere with equality only at the origin.

Answer 1

The Correct Option is D

Step 1: Find the stationary points.
A stationary point of \(f\) is a point where both partial derivatives vanish. Compute them from
\[ f(x_1,x_2)=2x_1^4+x_2^2+x_2x_1^2. \]
\[ \frac{\partial f}{\partial x_1}=8x_1^3+2x_1x_2=2x_1(4x_1^2+x_2) \]
\[ \frac{\partial f}{\partial x_2}=2x_2+x_1^2 \]

Step 2: Solve the two equations together.
From \(\partial f/\partial x_2=0\) we get \(x_2=-x_1^2/2\). Put this into \(\partial f/\partial x_1=0\):
\[ 2x_1\left(4x_1^2-\frac{x_1^2}{2}\right)=2x_1\cdot\frac{7x_1^2}{2}=7x_1^3=0 \]
This forces \(x_1=0\), and then \(x_2=-x_1^2/2=0\). So the only stationary point of \(f\) is \((0,0)\). There is no second one, so statement (A) is FALSE.

Step 3: Rewrite \(f\) as a sum of squares.
Treat \(f\) as a quadratic in \(x_2\) and complete the square:
\[ f(x_1,x_2)=x_2^2+x_1^2x_2+2x_1^4=\left(x_2+\frac{x_1^2}{2}\right)^2-\frac{x_1^4}{4}+2x_1^4 \]
\[ =\left(x_2+\frac{x_1^2}{2}\right)^2+\frac{7}{4}x_1^4 \]

Step 4: Read off the sign of \(f\).
Both terms on the right are squares, so \(f(x_1,x_2)\geq 0\) for every \((x_1,x_2)\), and \(f(x_1,x_2)=0\) only when \(x_2+\dfrac{x_1^2}{2}=0\) and \(x_1^4=0\) hold together, which forces \(x_1=0\) and then \(x_2=0\). So \(f(0,0)=0\) is the smallest value \(f\) ever takes, and no other point matches it.

Step 5: Classify \((0,0)\).
Since \(f(x_1,x_2)\geq f(0,0)\) for all \((x_1,x_2)\), with equality only at the origin, the point \((0,0)\) is a strict global minimum of \(f\), and hence also a local minimum. This rules out statement (B) (saddle point) and statement (C) (local maximum). The Hessian at the origin is
\[ H(0,0)=\begin{pmatrix}0 & 0\\ 0 & 2\end{pmatrix}, \]
which is only positive semi-definite (its determinant is \(0\)), so the second derivative test alone cannot decide the nature of \((0,0)\). The sum of squares form found in Step 3 settles the question directly.

Final Answer:
\((0,0)\) is a local minimum of \(f\), and it is the only stationary point. \[ \boxed{(0,0)\text{ is a local minimum of }f} \]