inflection f(x)=-x^3+3x^2-1 — step-by-step solution

1) For

- x^{3} + 3 x^{2} - 1

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

⇥1.1) Let's differentiate:

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

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

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

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

⇥

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

\frac{d}{d x} (3 x^{2})

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

\frac{d}{d x} (3 x^{2})

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

⇥

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

⇥ⓘ Multiplications are done before additions and subtractions:

3 \cdot 2

6

⇥

- 3 x^{2} + 6 x

⇥

6 x - 3 x^{2}

⇥You can write the expression in factored form:

x \cdot (6 - 3 x)

⇥Let's factorize in

ℝ

(6 - 3 x)

⇥We factor out

3

, the greatest common divisor of the coefficients:

6 - 3 x

3 \cdot (2 - x)

⇥Factor found:

2 - x

The first derivative is

3 x \cdot (2 - x)

⇥1.2) Let's differentiate:

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

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

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

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

⇥

3 \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)

⇥1.3) Let's perform the differentiation

u1

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

⇥ⓘ The derivative of x is 1:

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

1

⇥

1

⇥1.4) 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

⇥1.5) We 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

⇥1.6) Then you can continue the derivation:

⇥ⅰ

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

2 - 2 x

⇥

3 \cdot (2 - 2 x)

⇥ⓘ Expand by multiplying each term inside the parentheses:

3 \cdot (2 - 2 x)

6 - 6 x

⇥

6 - 6 x

The second derivative is

6 - 6 x

⇥1.7) Let's solve in

ℝ

6 - 6 x = 0

⇥ⓘ We add

6 x

to both sides

⇥

6 = 6 x

⇥ⓘ Both sides are divided by

6

⇥

x = \frac{6}{6}

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

6

, their greatest common divisor:

\frac{6}{6}

1

⇥Solution obtained:

x = 1

⇥1.8) Let's perform the differentiation:

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

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

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

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

⇥

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

⇥ⓘ Adding 0 changes nothing:

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

\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 = 1

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

Near

1

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

⇥

- 1^{3} + 3 \cdot 1^{2} - 1

⇥

- 1 + 3 - 1

⇥ⓘ Start from

-1

and move forward

3

on the number line, going past zero:

- 1 + 3

2

⇥

2 - 1

⇥ⓘ Start from

2

and move back

1

on the number line:

2 - 1

1

⇥

1

The y-coordinate is f(

1

) =

1

▶Inflection points:

{(1, 1)}

🖼/draw?f1=-x%5E3+%2B+3*x%5E2+-+1&v=x&b=-1.0&e=3.0&min=-1.4&max=3.4&points=1%2C1