Find the Inflection Points y=3x^3-9x-4

$y = 3 x^{3} - 9 x - 4$

Write $y = 3 x^{3} - 9 x - 4$ as a function.

$f (x) = 3 x^{3} - 9 x - 4$

Find the second derivative.

Find the first derivative.

By the Sum Rule, the derivative of $3 x^{3} - 9 x - 4$ with respect to $x$ is $\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [- 9 x] + \frac{d}{d x} [- 4]$ .

$f' (x) = \frac{d}{d x} (3 x^{3}) + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (- 4)$

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

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

$f' (x) = 3 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (- 4)$

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

$f' (x) = 3 (3 x^{2}) + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (- 4)$

Multiply $3$ by $3$ .

$f' (x) = 9 x^{2} + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (- 4)$

Evaluate $\frac{d}{d x} [- 9 x]$ .

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x$ with respect to $x$ is $- 9 \frac{d}{d x} [x]$ .

$f' (x) = 9 x^{2} - 9 \frac{d}{d x} x + \frac{d}{d x} (- 4)$

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

$f' (x) = 9 x^{2} - 9 \cdot 1 + \frac{d}{d x} (- 4)$

Multiply $- 9$ by $1$ .

$f' (x) = 9 x^{2} - 9 + \frac{d}{d x} (- 4)$

Differentiate using the Constant Rule.

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$f' (x) = 9 x^{2} - 9 + 0$

Add $9 x^{2} - 9$ and $0$ .

$f' (x) = 9 x^{2} - 9$

Find the second derivative.

By the Sum Rule, the derivative of $9 x^{2} - 9$ with respect to $x$ is $\frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [- 9]$ .

$f'' (x) = \frac{d}{d x} (9 x^{2}) + \frac{d}{d x} (- 9)$

Evaluate $\frac{d}{d x} [9 x^{2}]$ .

Since $9$ is constant with respect to $x$ , the derivative of $9 x^{2}$ with respect to $x$ is $9 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 9 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 9)$

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

$f'' (x) = 9 (2 x) + \frac{d}{d x} (- 9)$

Multiply $2$ by $9$ .

$f'' (x) = 18 x + \frac{d}{d x} (- 9)$

Differentiate using the Constant Rule.

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

$f'' (x) = 18 x + 0$

Add $18 x$ and $0$ .

$f'' (x) = 18 x$

The second derivative of $f (x)$ with respect to $x$ is $18 x$ .

$18 x$

Set the second derivative equal to $0$ then solve the equation $18 x = 0$ .

Set the second derivative equal to $0$ .

$18 x = 0$

Divide each term in $18 x = 0$ by $18$ and simplify.

Divide each term in $18 x = 0$ by $18$ .

$\frac{18 x}{18} = \frac{0}{18}$

Simplify the left side.

Cancel the common factor of $18$ .

Cancel the common factor.

$\frac{18 x}{18} = \frac{0}{18}$

Divide $x$ by $1$ .

$x = \frac{0}{18}$

Simplify the right side.

Divide $0$ by $18$ .

$x = 0$

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

Substitute $0$ in $f (x) = 3 x^{3} - 9 x - 4$ to find the value of $y$ .

Replace the variable $x$ with $0$ in the expression.

$f (0) = 3 {(0)}^{3} - 9 \cdot 0 - 4$

Simplify the result.

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$f (0) = 3 \cdot 0 - 9 \cdot 0 - 4$

Multiply $3$ by $0$ .

$f (0) = 0 - 9 \cdot 0 - 4$

Multiply $- 9$ by $0$ .

$f (0) = 0 + 0 - 4$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$f (0) = 0 - 4$

Subtract $4$ from $0$ .

$f (0) = - 4$

The final answer is $- 4$ .

$- 4$

The point found by substituting $0$ in $f (x) = 3 x^{3} - 9 x - 4$ is $(0, - 4)$ . This point can be an inflection point.

$(0, - 4)$

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

$(- \infty, 0) \cup (0, \infty)$

Substitute a value from the interval $(- \infty, 0)$ into the second derivative to determine if it is increasing or decreasing.

Replace the variable $x$ with $- 0.1$ in the expression.

$f'' (- 0.1) = 18 (- 0.1)$

Simplify the result.

Multiply $18$ by $- 0.1$ .

$f'' (- 0.1) = - 1.8$

The final answer is $- 1.8$ .

$- 1.8$

At $- 0.1$ , the second derivative is $- 1.8$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 0)$

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

Substitute a value from the interval $(0, \infty)$ into the second derivative to determine if it is increasing or decreasing.

Replace the variable $x$ with $0.1$ in the expression.

$f'' (0.1) = 18 (0.1)$

Simplify the result.

Multiply $18$ by $0.1$ .

$f'' (0.1) = 1.8$

The final answer is $1.8$ .

$1.8$

At $0.1$ , the second derivative is $1.8$ . Since this is positive, the second derivative is increasing on the interval $(0, \infty)$ .

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

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 $(0, - 4)$ .

$(0, - 4)$

Find the Inflection Points y=3x^3-9x-4

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