extreme f(x)=x^2e^x-4 — step-by-step solution

1) Find the strict local extrema of

x^{2} e^{x} - 4

: they are the points of the domain where its derivative changes sign

⇥Let's study the variations of

f (x)

: first find where its derivative

f^{'} (x) = e^{x} x \cdot (x + 2)

is zero

⇥Let's solve in

ℝ

e^{x} x \cdot (x + 2) = 0

⇥A product is zero if and only if one of its factors is zero: we solve each factor equal to 0

⇥⇥1.1) Let's solve in

ℝ

e^{x} = 0

⇥⇥In ℝ, the exponential function is strictly positive:

e^{x} > 0

, so

e^{x} = 0

has no solution

⇥⇥Solution:

x \in \emptyset

⇥⇥1.2) Let's solve in

ℝ

x = 0

⇥⇥Solution obtained:

x = 0

⇥⇥1.3) Let's solve in

ℝ

x + 2 = 0

⇥⇥Solution found:

x = -2

⇥

f^{'} (x)

is zero at

x = {-2, 0}

⇥Built from usual functions,

f^{'} (x)

is continuous where it is defined: on each interval where it is never zero, it keeps the sign of its value at any point

⇥

f^{'} (-3) = \frac{3}{e^{3}}

> 0:

f

is increasing on

(-∞, -2)

⇥

f^{'} (-1) = - \frac{1}{e}

< 0:

f

is decreasing on

(-2, 0)

⇥

f^{'} (1) = e \cdot 3

> 0:

f

is increasing on

(0, +∞)

⇥Let's calculate the values of

f

at the bounds of the table, or its limits at the infinite or excluded bounds

⇥As

x

tends to

-∞

f (x)

tends to

-4

⇥

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

⇥

f (0) = 0^{2} \cdot e^{0} - 4 = -4

⇥As

x

tends to

+∞

f (x)

tends to

+∞

⇥Let's draw up the variation table:

⇥

$x$	$-∞$		$-2$		$0$		$+∞$
$f^{'} (x)$		+	$0$	-	$0$	+
$f (x)$	$-4$	↗	$\frac{4}{e^{2}} - 4$	↘	$-4$	↗	$+∞$

⇥So

f

is increasing on

(-∞, -2]

, decreasing on

[-2, 0]

and increasing on

[0, +∞)

x = -2

f

' changes from + to -:

f

has a strict local maximum,

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

x = 0

f

' changes from - to +:

f

has a strict local minimum,

f (0) = -4

▶Strict local extrema:

{(-2, \frac{4}{e^{2}} - 4), (0, -4)}