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Popular >

extreme f(x)=(x^2-4)/(e^{x^4-1)}

Solution

extreme points

f (x) = \frac{x ^{2} - 4}{e ^{x^{4} - 1}}

Solution

Maximum - \frac{4 + 3 2}{2}, \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}, Minimum (0, - 4 e), Maximum \frac{4 + 3 2}{2}, \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}

Solution steps

f^{'} (x) = 2 e^{- x^{4} + 1} x (1 - 2 x^{2} (x^{2} - 4))

Find intervals:

Decreasing

: - \frac{4 + 3 2}{2} < x < 0,

Increasing

: 0 < x < \frac{4 + 3 2}{2}

Plug

x = - \frac{4 + 3 2}{2}

into

\frac{x ^{2} - 4}{e ^{x^{4} - 1}} : \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}

Maximum - \frac{4 + 3 2}{2}, \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}

Plug

x = 0

into

\frac{x ^{2} - 4}{e ^{x^{4} - 1}} : - 4 e

Minimum (0, - 4 e)

Plug

x = \frac{4 + 3 2}{2}

into

\frac{x ^{2} - 4}{e ^{x^{4} - 1}} : \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}

Maximum \frac{4 + 3 2}{2}, \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}

Maximum - \frac{4 + 3 2}{2}, \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}, Minimum (0, - 4 e), Maximum \frac{4 + 3 2}{2}, \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}

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Frequently Asked Questions (FAQ)

What is the extreme f(x)=(x^2-4)/(e^{x^4-1)} ?
The extreme f(x)=(x^2-4)/(e^{x^4-1)} is