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Critical points of -2x·e^(-x^2)

- 2 x e^{- x^{2}}

Step by step

  1. f(x) = - 2 x e^{- x^{2}}

    Critical points are where f′(x) = 0 or is undefined.

  2. f'(x) = 4 x^{2} e^{- x^{2}} - 2 e^{- x^{2}}

    Differentiate.

  3. x = - \frac{\sqrt{2}}{2}, x = \frac{\sqrt{2}}{2}

    Solve f′(x) = 0.

  4. f''(x) = - 8 x^{3} e^{- x^{2}} + 12 x e^{- x^{2}}

    Second-derivative test: f″ < 0 → maximum, f″ > 0 → minimum.

  5. f''(- \frac{\sqrt{2}}{2}) = - \frac{4 \sqrt{2}}{e^{\frac{1}{2}}} \Rightarrow (- \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{e^{\frac{1}{2}}}) \text{ is a local maximum}

  6. f''(\frac{\sqrt{2}}{2}) = \frac{4 \sqrt{2}}{e^{\frac{1}{2}}} \Rightarrow (\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{e^{\frac{1}{2}}}) \text{ is a local minimum}

Reveal the answer
(- \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{e^{\frac{1}{2}}})\ \text{local maximum},\; (\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{e^{\frac{1}{2}}})\ \text{local minimum}