What is the extreme f(x)=(x^2-x+1)e^x ?

The extreme f(x)=(x^2-x+1)e^x is Maximum(-1, 3/e),Minimum(0,1)

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extreme f(x)=(x^2-x+1)e^x

extreme points

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

Maximum (- 1, \frac{3}{e}), Minimum (0, 1)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 1

into

(x^{2} - x + 1) e^{x} : \frac{3}{e}

Maximum (- 1, \frac{3}{e})

Plug

x = 0

into

(x^{2} - x + 1) e^{x} : 1

Minimum (0, 1)

Maximum (- 1, \frac{3}{e}), Minimum (0, 1)

f (x) = - 2 x^{2} - 3

f (x) = \frac{x ^{3} - 1}{x ^{2} - 1}

f (x) = 3 x^{2} + 3

(0, 1000), (180, 100)

y = \frac{2 x ^{2} - 1}{4}

What is the extreme f(x)=(x^2-x+1)e^x ?
The extreme f(x)=(x^2-x+1)e^x is Maximum(-1, 3/e),Minimum(0,1)