inflection f(x)=x^3-6x^2+7 — step-by-step solution

1) For

x^{3} - 6 x^{2} + 7

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

⇥1.1) Let's differentiate:

\frac{d}{d x} (x^{3} - 6 x^{2} + 7)

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

\frac{d}{d x} (x^{3} - 6 x^{2} + 7)

\frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 6 x^{2}) + 0

⇥

\frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 6 x^{2}) + 0

⇥ⓘ The derivative of x^n is n*x^(n-1):

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

3 x^{2}

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

2 x

■Adding 0 changes nothing:

\frac{d}{d x} (- 6 x^{2}) + 0

\frac{d}{d x} (- 6 x^{2})

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

\frac{d}{d x} (- 6 x^{2})

-6 \cdot \frac{d}{d x} (x^{2})

⇥

3 x^{2} - 6 \cdot 2 x

⇥ⓘ Multiplications are done before additions and subtractions:

6 \cdot 2

12

⇥

3 x^{2} - 12 x

⇥You can factorize this expression into:

x \cdot (3 x - 12)

⇥Let's factor in

ℝ

(3 x - 12)

⇥We factor out

3

, the greatest common divisor of the coefficients:

3 x - 12

3 \cdot (x - 4)

⇥Factor identified:

x - 4

The first derivative is

3 x \cdot (x - 4)

⇥1.2) Let's differentiate:

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

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

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

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

⇥

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

⇥For

u1

x

and

v2

x - 4

, use the differentiation rule:

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

⇥1.3) Let's differentiate

u1

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

⇥ⓘ The derivative of x is 1:

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

1

⇥

1

⇥1.4) Let's perform the differentiation

v2

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

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

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

1 + 0

⇥

1 + 0

⇥ⓘ Adding 0 changes nothing:

1 + 0

1

⇥

1

⇥1.5) We evaluate the resulting derivative (

u1 \cdot v2

)':

1 \cdot (x - 4) + x \cdot 1

⇥

x - 4 + x

⇥ⓘ Like terms are grouped by adding their coefficients:

x + x

2 x

⇥

2 x - 4

⇥1.6) You can then resume the differentiation:

⇥ⅰ

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

2 x - 4

⇥

3 \cdot (2 x - 4)

⇥ⓘ Expand by multiplying each term inside the parentheses:

3 \cdot (2 x - 4)

6 x - 12

⇥

6 x - 12

The second derivative is

6 x - 12

⇥1.7) Let's solve in

ℝ

6 x - 12 = 0

⇥ⓘ We add

12

to both sides

⇥

6 x = 12

⇥ⓘ Both sides are divided by

6

⇥

x = \frac{12}{6}

⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

6

, their greatest common divisor:

\frac{12}{6}

2

⇥Solution found:

x = 2

⇥1.8) Let's differentiate:

\frac{d}{d x} (6 x - 12)

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

\frac{d}{d x} (6 x - 12)

\frac{d}{d x} (6 x) + 0

⇥

\frac{d}{d x} (6 x) + 0

⇥ⓘ Adding 0 changes nothing:

\frac{d}{d x} (6 x) + 0

\frac{d}{d x} (6 x)

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

\frac{d}{d x} (6 x)

6

⇥

6

x = 2

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

Near

2

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

⇥

2^{3} - 6 \cdot 2^{2} + 7

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

2^{3}

2 \cdot 2 \cdot 2

8

2^{2}

2 \cdot 2

4

⇥

8 - 6 \cdot 4 + 7

⇥ⓘ Multiplications are done before additions and subtractions:

6 \cdot 4

24

⇥

8 - 24 + 7

⇥ⓘ Start from

8

and move back

24

on the number line, going below zero:

8 - 24

-16

⇥

- 16 + 7

⇥ⓘ Start from

-16

and move forward

7

on the number line, getting closer to zero:

- 16 + 7

-9

⇥

-9

The y-coordinate is f(

2

) =

-9

▶Inflection points:

{(2, -9)}

🖼/draw?f1=x%5E3+-+6*x%5E2+%2B+7&v=x&b=0.0&e=4.0&min=-28.2&max=10.2&points=2%2C-9