Find the Inflection Points (x^3)/3-x^2-3x

$\frac{x^{3}}{3} - x^{2} - 3 x$

Step 1

Write $\frac{x^{3}}{3} - x^{2} - 3 x$ as a function.

$f (x) = \frac{x^{3}}{3} - x^{2} - 3 x$

Step 2

Find the second derivative.

Find the first derivative.

By the Sum Rule, the derivative of $\frac{x^{3}}{3} - x^{2} - 3 x$ with respect to $x$ is $\frac{d}{d x} [\frac{x^{3}}{3}] + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 3 x]$ .

$f' (x) = \frac{d}{d x} (\frac{x^{3}}{3}) + \frac{d}{d x} (- x^{2}) + \frac{d}{d x} (- 3 x)$

Evaluate $\frac{d}{d x} [\frac{x^{3}}{3}]$ .

Since $\frac{1}{3}$ is constant with respect to $x$ , the derivative of $\frac{x^{3}}{3}$ with respect to $x$ is $\frac{1}{3} \frac{d}{d x} [x^{3}]$ .

$f' (x) = \frac{1}{3} \cdot \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- x^{2}) + \frac{d}{d x} (- 3 x)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f' (x) = \frac{1}{3} \cdot (3 x^{2}) + \frac{d}{d x} (- x^{2}) + \frac{d}{d x} (- 3 x)$

Combine $3$ and $\frac{1}{3}$ .

$f' (x) = \frac{3}{3} x^{2} + \frac{d}{d x} (- x^{2}) + \frac{d}{d x} (- 3 x)$

Combine $\frac{3}{3}$ and $x^{2}$ .

$f' (x) = \frac{3 x^{2}}{3} + \frac{d}{d x} (- x^{2}) + \frac{d}{d x} (- 3 x)$

Cancel the common factor of $3$ .

Cancel the common factor.

$f' (x) = \frac{3 x^{2}}{3} + \frac{d}{d x} (- x^{2}) + \frac{d}{d x} (- 3 x)$

Divide $x^{2}$ by $1$ .

$f' (x) = x^{2} + \frac{d}{d x} (- x^{2}) + \frac{d}{d x} (- 3 x)$

Evaluate $\frac{d}{d x} [- x^{2}]$ .

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$f' (x) = x^{2} - \frac{d}{d x} x^{2} + \frac{d}{d x} (- 3 x)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f' (x) = x^{2} - (2 x) + \frac{d}{d x} (- 3 x)$

Multiply $2$ by $- 1$ .

$f' (x) = x^{2} - 2 x + \frac{d}{d x} (- 3 x)$

Evaluate $\frac{d}{d x} [- 3 x]$ .

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x$ with respect to $x$ is $- 3 \frac{d}{d x} [x]$ .

$f' (x) = x^{2} - 2 x - 3 \frac{d}{d x} x$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f' (x) = x^{2} - 2 x - 3 \cdot 1$

Multiply $- 3$ by $1$ .

$f' (x) = x^{2} - 2 x - 3$

Find the second derivative.

Differentiate.

By the Sum Rule, the derivative of $x^{2} - 2 x - 3$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [- 3]$ .

$f'' (x) = \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (- 3)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = 2 x + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (- 3)$

Evaluate $\frac{d}{d x} [- 2 x]$ .

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

$f'' (x) = 2 x - 2 \frac{d}{d x} x + \frac{d}{d x} (- 3)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = 2 x - 2 \cdot 1 + \frac{d}{d x} (- 3)$

Multiply $- 2$ by $1$ .

$f'' (x) = 2 x - 2 + \frac{d}{d x} (- 3)$

Differentiate using the Constant Rule.

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$f'' (x) = 2 x - 2 + 0$

Add $2 x - 2$ and $0$ .

$f'' (x) = 2 x - 2$

The second derivative of $f (x)$ with respect to $x$ is $2 x - 2$ .

$2 x - 2$

Step 3

Set the second derivative equal to $0$ then solve the equation $2 x - 2 = 0$ .

Set the second derivative equal to $0$ .

$2 x - 2 = 0$

Add $2$ to both sides of the equation.

$2 x = 2$

Divide each term in $2 x = 2$ by $2$ and simplify.

Divide each term in $2 x = 2$ by $2$ .

$\frac{2 x}{2} = \frac{2}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{2}{2}$

Divide $x$ by $1$ .

$x = \frac{2}{2}$

Simplify the right side.

Divide $2$ by $2$ .

$x = 1$

Step 4

Find the points where the second derivative is $0$ .

Substitute $1$ in $f (x) = \frac{x^{3}}{3} - x^{2} - 3 x$ to find the value of $y$ .

Replace the variable $x$ with $1$ in the expression.

$f (1) = \frac{{(1)}^{3}}{3} - {(1)}^{2} - 3 \cdot 1$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = \frac{1}{3} - {(1)}^{2} - 3 \cdot 1$

One to any power is one.

$f (1) = \frac{1}{3} - 1 \cdot 1 - 3 \cdot 1$

Multiply $- 1$ by $1$ .

$f (1) = \frac{1}{3} - 1 - 3 \cdot 1$

Multiply $- 3$ by $1$ .

$f (1) = \frac{1}{3} - 1 - 3$

Find the common denominator.

Write $- 1$ as a fraction with denominator $1$ .

$f (1) = \frac{1}{3} + \frac{- 1}{1} - 3$

Multiply $\frac{- 1}{1}$ by $\frac{3}{3}$ .

$f (1) = \frac{1}{3} + \frac{- 1}{1} \cdot \frac{3}{3} - 3$

Multiply $\frac{- 1}{1}$ by $\frac{3}{3}$ .

$f (1) = \frac{1}{3} + \frac{- 1 \cdot 3}{3} - 3$

Write $- 3$ as a fraction with denominator $1$ .

$f (1) = \frac{1}{3} + \frac{- 1 \cdot 3}{3} + \frac{- 3}{1}$

Multiply $\frac{- 3}{1}$ by $\frac{3}{3}$ .

$f (1) = \frac{1}{3} + \frac{- 1 \cdot 3}{3} + \frac{- 3}{1} \cdot \frac{3}{3}$

Multiply $\frac{- 3}{1}$ by $\frac{3}{3}$ .

$f (1) = \frac{1}{3} + \frac{- 1 \cdot 3}{3} + \frac{- 3 \cdot 3}{3}$

Combine the numerators over the common denominator.

$f (1) = \frac{1 - 1 \cdot 3 - 3 \cdot 3}{3}$

Simplify each term.

Multiply $- 1$ by $3$ .

$f (1) = \frac{1 - 3 - 3 \cdot 3}{3}$

Multiply $- 3$ by $3$ .

$f (1) = \frac{1 - 3 - 9}{3}$

Simplify the expression.

Subtract $3$ from $1$ .

$f (1) = \frac{- 2 - 9}{3}$

Subtract $9$ from $- 2$ .

$f (1) = \frac{- 11}{3}$

Move the negative in front of the fraction.

$f (1) = - \frac{11}{3}$

The final answer is $- \frac{11}{3}$ .

$- \frac{11}{3}$

The point found by substituting $1$ in $f (x) = \frac{x^{3}}{3} - x^{2} - 3 x$ is $(1, - \frac{11}{3})$ . This point can be an inflection point.

$(1, - \frac{11}{3})$

Step 5

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, 1) \cup (1, \infty)$

Step 6

Substitute a value from the interval $(- \infty, 1)$ into the second derivative to determine if it is increasing or decreasing.

Replace the variable $x$ with $0.9$ in the expression.

$f'' (0.9) = 2 (0.9) - 2$

Simplify the result.

Multiply $2$ by $0.9$ .

$f'' (0.9) = 1.8 - 2$

Subtract $2$ from $1.8$ .

$f'' (0.9) = - 0.2$

The final answer is $- 0.2$ .

$- 0.2$

At $0.9$ , the second derivative is $- 0.2$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 1)$

Decreasing on $(- \infty, 1)$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(1, \infty)$ into the second derivative to determine if it is increasing or decreasing.

Replace the variable $x$ with $1.1$ in the expression.

$f'' (1.1) = 2 (1.1) - 2$

Simplify the result.

Multiply $2$ by $1.1$ .

$f'' (1.1) = 2.2 - 2$

Subtract $2$ from $2.2$ .

$f'' (1.1) = 0.2$

The final answer is $0.2$ .

$0.2$

At $1.1$ , the second derivative is $0.2$ . Since this is positive, the second derivative is increasing on the interval $(1, \infty)$ .

Increasing on $(1, \infty)$ since $f'' (x) > 0$

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(1, - \frac{11}{3})$ .

$(1, - \frac{11}{3})$

Step 9

Find the Inflection Points (x^3)/3-x^2-3x

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