extreme 60x^2-20x^3 — step-by-step solution

1) Find the strict local extrema of

60 x^{2} - 20 x^{3}

on ℝ: the derivative must be zero and change sign

⇥1.1) Let's differentiate:

\frac{d}{d x} (60 x^{2} - 20 x^{3})

⇥ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (60 x^{2} - 20 x^{3})

\frac{d}{d x} (60 x^{2}) + \frac{d}{d x} (- 20 x^{3})

⇥

\frac{d}{d x} (60 x^{2}) + \frac{d}{d x} (- 20 x^{3})

⇥ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (60 x^{2})

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

\frac{d}{d x} (- 20 x^{3})

-20 \cdot \frac{d}{d x} (x^{3})

■The derivative of x^n is n*x^(n-1):

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

2 x

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

3 x^{2}

⇥

60 \cdot 2 x - 20 \cdot 3 x^{2}

⇥ⓘ Multiplications are done before additions and subtractions:

60 \cdot 2

120

20 \cdot 3

60

⇥

120 x - 60 x^{2}

⇥This expression can be factorized into:

x \cdot (120 - 60 x)

⇥Let's factor in

ℝ

(120 - 60 x)

⇥We factor out

60

, the greatest common divisor of the coefficients:

120 - 60 x

60 \cdot (2 - x)

⇥Factor identified:

2 - x

⇥The first derivative is

60 x \cdot (2 - x)

⇥1.2) Let's solve in

ℝ

60 x \cdot (2 - x) = 0

⇥We simplify the expression by

60

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

⇥⇥Let's solve in

ℝ

x = 0

⇥⇥Solution found:

x = 0

⇥⇥Let's solve in

ℝ

2 - x = 0

⇥⇥Solution obtained:

x = 2

⇥1.3) Let's differentiate:

\frac{d}{d x} (60 x \cdot (2 - x))

⇥ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (60 x \cdot (2 - x))

60 \cdot \frac{d}{d x} (x \cdot (2 - x))

⇥

60 \cdot \frac{d}{d x} (x \cdot (2 - x))

⇥We differentiate

u1

x

and

v2

2 - x

using:

\frac{d}{d x} (u1 \cdot v2) = \frac{d}{d x} (u1) \cdot v2 + u1 \cdot \frac{d}{d x} (v2)

⇥1.4) Let's perform the differentiation

u1

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

⇥ⓘ The derivative of x is 1:

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

1

⇥

1

⇥1.5) Let's differentiate

v2

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

⇥ⓘ The derivative of a sum is the sum of the derivatives:

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

0 + \frac{d}{d x} (- x)

⇥

0 + \frac{d}{d x} (- x)

⇥ⓘ Adding 0 changes nothing:

0 + \frac{d}{d x} (- x)

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

■The derivative of x is 1:

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

-1

⇥

-1

⇥1.6) Let's evaluate the resulting derivative (

u1 \cdot v2

)':

1 \cdot (2 - x) + x \cdot (-1)

⇥ⓘ Multiplying by -1 gives the opposite:

x \cdot (-1)

- x

⇥

2 - x - x

⇥ⓘ Like terms are grouped by adding their coefficients:

- x - x

-2 x

⇥

2 - 2 x

⇥1.7) You can then resume the differentiation:

⇥ⅰ

\frac{d}{d x} (x \cdot (2 - x))

2 - 2 x

⇥

60 \cdot (2 - 2 x)

⇥ⓘ Expand by multiplying each term inside the parentheses:

60 \cdot (2 - 2 x)

120 - 120 x

⇥

120 - 120 x

⇥We factor out

120

, the greatest common divisor of the coefficients:

120 - 120 x

120 \cdot (1 - x)

⇥Factor found:

1 - x

⇥The second derivative is

120 \cdot (1 - x)

2) At

x = 0

, calculate

f^{''} (0) = 120 \cdot (1 + 0) = 120

x = 0

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

⇥

60 \cdot 0^{2} - 20 \cdot 0^{3}

⇥ⓘ A product with a zero factor is zero:

60 \cdot 0

0

-20 \cdot 0

0

⇥

0 + 0

⇥ⓘ Adding 0 changes nothing:

0 + 0

0

⇥

0

The y-coordinate is f(

0

) =

0

3) At

x = 2

, calculate

f^{''} (2) = 120 \cdot (1 - 2) = -120

x = 2

, the second derivative is negative: this is a strict local maximum

⇥

60 \cdot 2^{2} - 20 \cdot 2^{3}

⇥ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

2^{2}

2 \cdot 2

4

2^{3}

2 \cdot 2 \cdot 2

8

⇥

60 \cdot 4 - 20 \cdot 8

⇥ⓘ Multiplications are done before additions and subtractions:

60 \cdot 4

240

20 \cdot 8

160

⇥

240 - 160

⇥ⓘ Start from

240

and move back

160

on the number line:

240 - 160

80

⇥

80

The y-coordinate is f(

2

) =

80

This odd-degree polynomial is unbounded above and below on ℝ: it has no global extremum

▶Strict local extrema:

{(0, 0), (2, 80)}