Exercise · ZW-SECOND-DERIVATIVE-TEST-004 · ★☆☆ Basic

The function \(f(x,y)=x^3-3x+y^2\) has critical points at \((1,0)\) and \((-1,0)\). What does the test say at \((-1,0)\)?

Choose one option.

Show answer
  • C. Saddle point

Explanation

At \((-1,0)\): \(f_{xx}=6x=-6\), \(f_{yy}=2\), \(f_{xy}=0\), so \(D=-12<0\): saddle point. (At \((1,0)\) there would be a minimum.)

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Review the concept: Second derivative test

The second derivative test classifies a critical point of \(f\) as a minimum, maximum or saddle according to the sign of the discriminant \(D\) and of \(f_{xx}\) at that point.

\[D=f_{xx}f_{yy}-\left(f_{xy}\right)^{2}\]

More practice problems: second derivative test

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