What is the critical f(x)=1-e^{x^2+4x},-4<= x<= 0 ?

The critical f(x)=1-e^{x^2+4x},-4<= x<= 0 is x=-4,x=-2,x=0

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critical f(x)=1-e^{x^2+4x},-4<= x<= 0

critical points

f (x) = 1 - e^{x^{2} + 4 x}, - 4 \leq x \leq 0

x = - 4, x = - 2, x = 0

Solution steps

1 - e^{x^{2} + 4 x}, - 4 \leq x \leq 0

Find where

f^{'} (x)

is equal to zero or undefined

x = - 2

Identify critical points not in

f (x)

domain

Domain of

1 - e^{x^{2} + 4 x}, - 4 \leq x \leq 0 : - 4 \leq x \leq 0

All critical points are in domain

x = - 2

Evaluate interval endpoint

x = - 4 :

critical point

Evaluate interval endpoint

x = 0 :

critical point

x = - 4, x = - 2, x = 0

x^{3} - 3 x + 3 x y^{2}

\frac{4 x ^{5}}{5} - \frac{5 x ^{4}}{4} + 4 x - 1

f (x) = x x^{2} + 36

f (x) = e^{2 x}

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

What is the critical f(x)=1-e^{x^2+4x},-4<= x<= 0 ?
The critical f(x)=1-e^{x^2+4x},-4<= x<= 0 is x=-4,x=-2,x=0