extreme f(x)=x^4-4x^3+9 — step-by-step solution

⚠ 4x is read as the product 4*x: for a power, write 4^x

1) Find the strict local extrema of

x^{4} - 4 x^{3} + 9

on ℝ: the derivative must be zero and change sign

⇥1.1) Let's differentiate:

\frac{d}{d x} (x^{4} - 4 x^{3} + 9)

⇥

\frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 4 x^{3}) + 0

⇥ℹ

\frac{d}{d x} (x^{4})

4 x^{3}

‖

\frac{d}{d x} (- 4 x^{3})

-4 \cdot \frac{d}{d x} (x^{3})

‖

\frac{d}{d x} (x^{3})

3 x^{2}

⇥

4 x^{3} - 4 \cdot 3 x^{2}

⇥

4 x^{3} - 12 x^{2}

⇥This expression can be factorized into:

x^{2} \cdot (4 x - 12)

⇥1.2) Let's factor in

ℝ

(4 x - 12)

⇥This expression can be factorized into:

4 \cdot (x - 3)

⇥Factor found:

(x - 3)

⇥The first derivative is

4 x^{2} \cdot (x - 3)

⇥1.3) Let's solve in

ℝ

4 x^{2} \cdot (x - 3) = 0

⇥The expression can be simplified by:

4

⇥Let's solve in

ℝ

x^{2} = 0

⇥Solution obtained:

x = 0

⇥Let's solve in

ℝ

x - 3 = 0

⇥Solution found:

x = 3

⇥1.4) Let's perform the differentiation:

\frac{d}{d x} (4 x^{2} \cdot (x - 3))

⇥

4 \cdot \frac{d}{d x} (x^{2} \cdot (x - 3))

⇥For

U1

x^{2}

and

V2

x - 3

, use the product/quotient rule:

\frac{d}{d x} (U1 \cdot V2) = \frac{d}{d x} (U1) \cdot V2 + U1 \cdot \frac{d}{d x} (V2)

⇥1.5) Let's perform the differentiation

U1

\frac{d}{d x} (x^{2})

⇥

2 x

⇥1.6) Let's differentiate

V2

\frac{d}{d x} (x - 3)

⇥

1 + 0

⇥

1

⇥1.7) You can evaluate the resulted derivative (

U1 \cdot V2

)':

2 x \cdot (x - 3) + x^{2} \cdot 1

⇥

2 x \cdot (x - 3) + x^{2}

⇥

(2 x^{2} - 6 x) + x^{2}

⇥

3 x^{2} - 6 x

⇥1.8) Then you can continue the derivation:

⇥

4 \cdot (3 x^{2} - 6 x)

⇥

12 x^{2} - 24 x

⇥You can factorize this expression into:

x \cdot (12 x - 24)

⇥1.9) Let's factor in

ℝ

(12 x - 24)

⇥You can write the expression in factored form:

12 \cdot (x - 2)

⇥Factor identified:

(x - 2)

⇥The second derivative is

12 x \cdot (x - 2)

2) At

x = 0

, calculate

f'' (0) = 12 \cdot 0 \cdot (0 - 2) = 0

The first nonzero derivative at this point has order 3 and value

-24

x = 0

, the derivative has a zero of even order: it does not change sign, so this is not an extremum

3) At

x = 3

, calculate

f'' (3) = 12 \cdot 3 \cdot (3 - 2) = 36

x = 3

, the second derivative is positive: this is a strict local minimum

⇥

3^{4} - 4 \cdot 3^{3} + 9

⇥ℹ

3^{4}

81

‖

3^{3}

27

⇥

81 - 4 \cdot 27 + 9

⇥ℹ

- 4 \cdot 27

-108

‖

81 + 9

90

⇥

90 - 108

⇥

-18

The y-coordinate is f(

3

) =

-18

The polynomial tends to +∞ at both infinities: this unique local extremum is also global

▶Strict local extrema:

{(3, -18)}