extreme 4x(x^2-9) — step-by-step solution

Find the strict local extrema of

4 x \cdot (x^{2} - 9)

: 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) = 12 x^{2} - 36

is zero

⇥Let's solve in

ℝ

12 x^{2} - 36 = 0

⇥ⓘ Both sides are divided by

12

⇥

x^{2} = 3

⇥The two numbers whose square is

3

are

- \sqrt{3}

and

\sqrt{3}

⇥Solutions obtained:

x = {- \sqrt{3}, \sqrt{3}}

⇥

f^{'} (x)

is zero at

x = {- \sqrt{3}, \sqrt{3}}

⇥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^{'} (-2) = 12

> 0:

f

is increasing on

(-∞, - \sqrt{3})

⇥

f^{'} (0) = -36

< 0:

f

is decreasing on

(- \sqrt{3}, \sqrt{3})

⇥

f^{'} (2) = 12

> 0:

f

is increasing on

(\sqrt{3}, +∞)

⇥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

-∞

⇥

f (- \sqrt{3}) = 4 \cdot (- \sqrt{3}) \cdot ({(- \sqrt{3})}^{2} - 9) = 24 \cdot \sqrt{3}

⇥

f (\sqrt{3}) = 4 \cdot \sqrt{3} \cdot ({(\sqrt{3})}^{2} - 9) = - 24 \cdot \sqrt{3}

⇥As

x

tends to

+∞

f (x)

tends to

+∞

⇥Let's draw up the variation table:

⇥

⇥So

f

is increasing on

(-∞, - \sqrt{3}]

, decreasing on

[- \sqrt{3}, \sqrt{3}]

and increasing on

[\sqrt{3}, +∞)

x = - \sqrt{3}

f

' changes from + to -:

f

has a strict local maximum,

f (- \sqrt{3}) = 24 \cdot \sqrt{3}

x = \sqrt{3}

f

' changes from - to +:

f

has a strict local minimum,

f (\sqrt{3}) = - 24 \cdot \sqrt{3}

▶Strict local extrema:

{(- \sqrt{3}, 24 \cdot \sqrt{3}), (\sqrt{3}, - 24 \cdot \sqrt{3})}