inflection 3x^5-10x^3 — step-by-step solution

1) For

3 x^{5} - 10 x^{3}

on ℝ, find where the second derivative is zero and changes sign

⇥1.1) Let's differentiate:

\frac{d}{d x} (3 x^{5} - 10 x^{3})

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

\frac{d}{d x} (3 x^{5} - 10 x^{3})

\frac{d}{d x} (3 x^{5}) + \frac{d}{d x} (- 10 x^{3})

⇥

\frac{d}{d x} (3 x^{5}) + \frac{d}{d x} (- 10 x^{3})

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

\frac{d}{d x} (3 x^{5})

3 \cdot \frac{d}{d x} (x^{5})

\frac{d}{d x} (- 10 x^{3})

-10 \cdot \frac{d}{d x} (x^{3})

■The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (x^{5})

5 x^{4}

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

3 x^{2}

⇥

3 \cdot 5 x^{4} - 10 \cdot 3 x^{2}

⇥ⓘ Multiplications are done before additions and subtractions:

3 \cdot 5

15

10 \cdot 3

30

⇥

15 x^{4} - 30 x^{2}

The first derivative is

15 x^{4} - 30 x^{2}

⇥1.2) Let's perform the differentiation:

\frac{d}{d x} (15 x^{4} - 30 x^{2})

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

\frac{d}{d x} (15 x^{4} - 30 x^{2})

\frac{d}{d x} (15 x^{4}) + \frac{d}{d x} (- 30 x^{2})

⇥

\frac{d}{d x} (15 x^{4}) + \frac{d}{d x} (- 30 x^{2})

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

\frac{d}{d x} (15 x^{4})

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

\frac{d}{d x} (- 30 x^{2})

-30 \cdot \frac{d}{d x} (x^{2})

■The derivative of x^n is n*x^(n-1):

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

4 x^{3}

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

2 x

⇥

15 \cdot 4 x^{3} - 30 \cdot 2 x

⇥ⓘ Multiplications are done before additions and subtractions:

15 \cdot 4

(10 + 5) \cdot 4

10 \cdot 4 + 5 \cdot 4

40 + 20

60

30 \cdot 2

60

⇥

60 x^{3} - 60 x

The second derivative is

60 x^{3} - 60 x

⇥1.3) Let's solve in

ℝ

60 x^{3} - 60 x = 0

⇥You can write the expression:

60 x^{3} - 60 x

in factored form:

x \cdot (60 x^{2} - 60)

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

⇥⇥1.3.1) Let's solve in

ℝ

x = 0

⇥⇥Solution found:

x = 0

⇥⇥1.3.2) Let's solve in

ℝ

60 x^{2} - 60 = 0

⇥⇥ⓘ Both sides are divided by

60

⇥⇥

x^{2} = 1

⇥⇥The two numbers whose square is

1

are

-1

and

1

⇥⇥Solutions found:

x = {-1, 1}

⇥1.4) Let's perform the differentiation:

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

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

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

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

⇥

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

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

\frac{d}{d x} (60 x^{3})

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

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

-60

■The derivative of x^n is n*x^(n-1):

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

3 x^{2}

⇥

60 \cdot 3 x^{2} - 60

⇥ⓘ Multiplications are done before additions and subtractions:

60 \cdot 3

180

⇥

180 x^{2} - 60

x = -1

, the zero of the second derivative has order 1: this order is odd, so the concavity changes

Near

-1

, the second derivative is negative on the left and positive on the right: the curve changes from concave down to concave up

⇥

3 \cdot {(-1)}^{5} - 10 \cdot {(-1)}^{3}

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

{(-1)}^{5}

(-1) \cdot ... \cdot (-1)

, 5 times =

-1

{(-1)}^{3}

(-1) \cdot (-1) \cdot (-1)

-1

■Multiplying by -1 gives the opposite:

3 \cdot (-1)

-3

- 10 \cdot (-1)

10

⇥

- 3 + 10

⇥ⓘ Start from

-3

and move forward

10

on the number line, going past zero:

- 3 + 10

7

⇥

7

The y-coordinate is f(

-1

) =

7

⇥1.5) Let's perform the differentiation:

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

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

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

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

⇥

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

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

\frac{d}{d x} (60 x^{3})

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

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

-60

■The derivative of x^n is n*x^(n-1):

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

3 x^{2}

⇥

60 \cdot 3 x^{2} - 60

⇥ⓘ Multiplications are done before additions and subtractions:

60 \cdot 3

180

⇥

180 x^{2} - 60

x = 0

, the zero of the second derivative has order 1: this order is odd, so the concavity changes

Near

0

, the second derivative is positive on the left and negative on the right: the curve changes from concave up to concave down

⇥

3 \cdot 0^{5} - 10 \cdot 0^{3}

⇥ⓘ A product with a zero factor is zero:

3 \cdot 0

0

10 \cdot 0

0

⇥

0 + 0

⇥ⓘ Adding 0 changes nothing:

0 + 0

0

⇥

0

The y-coordinate is f(

0

) =

0

⇥1.6) Let's differentiate:

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

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

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

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

⇥

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

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

\frac{d}{d x} (60 x^{3})

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

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

-60

■The derivative of x^n is n*x^(n-1):

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

3 x^{2}

⇥

60 \cdot 3 x^{2} - 60

⇥ⓘ Multiplications are done before additions and subtractions:

60 \cdot 3

180

⇥

180 x^{2} - 60

x = 1

, the zero of the second derivative has order 1: this order is odd, so the concavity changes

Near

1

, the second derivative is negative on the left and positive on the right: the curve changes from concave down to concave up

⇥

3 \cdot 1^{5} - 10 \cdot 1^{3}

⇥

3 - 10

⇥ⓘ Start from

3

and move back

10

on the number line, going below zero:

3 - 10

-7

⇥

-7

The y-coordinate is f(

1

) =

-7

▶Inflection points:

{(-1, 7), (0, 0), (1, -7)}

🖼/draw?f1=3*x%5E5+-+10*x%5E3&v=x&b=-2.0&e=2.0&min=-19.2&max=19.2&points=-1%2C7%3B0%2C0%3B1%2C-7