Use calculus to find the maximum and minimum values of the function f(x)=e^((x^7)-x) in -1<x<0

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The function `f(x)=e^((x^7)-x)`

The extreme points of the function are obtained by solving the equation f'(x) = 0

f'(x) = `e^(x^7 - x)*(7x^6 - 1)`

`e^(x^7 - x)*(7x^6 - 1) = 0`

=> `7x^6 = 1`

=> `x^6 = 1/7`

The real root of the equation greater than -1 and less than 0 is at `x = -(1/7)^(1/6)` .

**The function `f(x)=e^((x^7)-x)` has a maximum at `x = -(1/7)^(1/6)` .**

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