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: Recall what stationary point and Hessian test mean.
A stationary point of a function of two variables is a point where both first partial derivatives vanish. Once we find such a point, the usual next step is the second derivative (Hessian) test: if the Hessian matrix of second partial derivatives is positive definite there, the point is a local minimum; if negative definite, a local maximum; if it has both a positive and a negative eigenvalue, it is a saddle point. When the Hessian is only semi-definite, meaning its determinant is zero, this test gives no answer and we must argue directly from the function.

Step 2: Find the stationary point a different way.
Set the first partial derivative to zero first: $\dfrac{\partial f}{\partial x_1}=2x_1(4x_1^2+x_2)=0$ gives $x_1=0$ or $x_2=-4x_1^2$.

Step 3: Feed each branch into the second equation.
The second equation is $\dfrac{\partial f}{\partial x_2}=2x_2+x_1^2=0$. If $x_1=0$, this gives $x_2=0$, so we get the point $(0,0)$. If instead $x_2=-4x_1^2$, substitute to get $2(-4x_1^2)+x_1^2=-7x_1^2=0$, so again $x_1=0$ and then $x_2=0$. Both branches collapse to the same point, so $(0,0)$ is the only stationary point and there cannot be a second distinct one. This already rules out option (A), which claims two distinct stationary points.

Step 4: Test positivity with a discriminant argument.
View $f$ as a quadratic in $x_2$ for a fixed $x_1$:
\[ f=x_2^2+x_1^2x_2+2x_1^4. \]
The coefficient of $x_2^2$ is $1>0$, so this quadratic opens upward in $x_2$. Its discriminant is
\[ D=(x_1^2)^2-4(1)(2x_1^4)=x_1^4-8x_1^4=-7x_1^4. \]

Step 5: Read the discriminant.
Since $x_1^4\geq 0$, we have $D=-7x_1^4\leq 0$ for every real $x_1$, with $D=0$ only at $x_1=0$. A upward quadratic with $D\leq 0$ never goes negative, so $f\geq 0$ for every $(x_1,x_2)$. When $x_1\neq 0$ the discriminant is strictly negative, so $f>0$ strictly, and when $x_1=0$ the quadratic reduces to $x_2^2\geq 0$, zero only at $x_2=0$.

Step 6: Cross check with a couple of test points.
Take $x_1=1,x_2=-2$: $f=2(1)+4+(-2)(1)=2+4-2=4>0$. Take $x_1=1,x_2=0$: $f=2(1)+0+0=2>0$. Take $x_1=0,x_2=1$: $f=0+1+0=1>0$. None of these ever beat $f(0,0)=0$, consistent with $(0,0)$ being the minimum.

Step 7: Conclude.
So $f(x_1,x_2)\geq 0=f(0,0)$ everywhere, and this value is achieved nowhere else. That makes $(0,0)$ a strict global minimum, so it is a local minimum, not a saddle point or a local maximum. \[ \boxed{(0,0)\text{ is a local minimum of }f} \]