Find the Concavity f(x)=x^3+3x^2-9x+5

$f (x) = x^{3} + 3 x^{2} - 9 x + 5$

Step 1

Find the $x$ values where the second derivative is equal to $0$ .

Find the second derivative.

Find the first derivative.

Differentiate.

By the Sum Rule, the derivative of $x^{3} + 3 x^{2} - 9 x + 5$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 9 x] + \frac{d}{d x} [5]$ .

$f' (x) = \frac{d}{d x} (x^{3}) + \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (5)$

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

$f' (x) = 3 x^{2} + \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (5)$

Evaluate $\frac{d}{d x} [3 x^{2}]$ .

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 3 x^{2} + 3 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (5)$

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

$f' (x) = 3 x^{2} + 3 (2 x) + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (5)$

Multiply $2$ by $3$ .

$f' (x) = 3 x^{2} + 6 x + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (5)$

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

$f' (x) = 3 x^{2} + 6 x - 9 \frac{d}{d x} x + \frac{d}{d x} (5)$

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

$f' (x) = 3 x^{2} + 6 x - 9 \cdot 1 + \frac{d}{d x} (5)$

Multiply $- 9$ by $1$ .

$f' (x) = 3 x^{2} + 6 x - 9 + \frac{d}{d x} (5)$

Differentiate using the Constant Rule.

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$f' (x) = 3 x^{2} + 6 x - 9 + 0$

Add $3 x^{2} + 6 x - 9$ and $0$ .

$f' (x) = 3 x^{2} + 6 x - 9$

Find the second derivative.

By the Sum Rule, the derivative of $3 x^{2} + 6 x - 9$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 9]$ .

$f'' (x) = \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 9)$

Evaluate $\frac{d}{d x} [3 x^{2}]$ .

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 3 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 9)$

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

$f'' (x) = 3 (2 x) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 9)$

Multiply $2$ by $3$ .

$f'' (x) = 6 x + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 9)$

Evaluate $\frac{d}{d x} [6 x]$ .

Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$f'' (x) = 6 x + 6 \frac{d}{d x} (x) + \frac{d}{d x} (- 9)$

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

$f'' (x) = 6 x + 6 \cdot 1 + \frac{d}{d x} (- 9)$

Multiply $6$ by $1$ .

$f'' (x) = 6 x + 6 + \frac{d}{d x} (- 9)$

Differentiate using the Constant Rule.

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

$f'' (x) = 6 x + 6 + 0$

Add $6 x + 6$ and $0$ .

$f'' (x) = 6 x + 6$

The second derivative of $f (x)$ with respect to $x$ is $6 x + 6$ .

$6 x + 6$

Set the second derivative equal to $0$ then solve the equation $6 x + 6 = 0$ .

Set the second derivative equal to $0$ .

$6 x + 6 = 0$

Subtract $6$ from both sides of the equation.

$6 x = - 6$

Divide each term in $6 x = - 6$ by $6$ and simplify.

Divide each term in $6 x = - 6$ by $6$ .

$\frac{6 x}{6} = \frac{- 6}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{- 6}{6}$

Divide $x$ by $1$ .

$x = \frac{- 6}{6}$

Simplify the right side.

Divide $- 6$ by $6$ .

$x = - 1$

Step 2

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3

Create intervals around the $x$ -values where the second derivative is zero or undefined.

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

Step 4

Substitute any number from the interval $(- \infty, - 1)$ into the second derivative and evaluate to determine the concavity.

Replace the variable $x$ with $- 4$ in the expression.

$f'' (- 4) = 6 (- 4) + 6$

Simplify the result.

Multiply $6$ by $- 4$ .

$f'' (- 4) = - 24 + 6$

Add $- 24$ and $6$ .

$f'' (- 4) = - 18$

The final answer is $- 18$ .

$- 18$

The graph is concave down on the interval $(- \infty, - 1)$ because $f'' (- 4)$ is negative.

Concave down on $(- \infty, - 1)$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(- 1, \infty)$ into the second derivative and evaluate to determine the concavity.

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

$f'' (0) = 6 (0) + 6$

Simplify the result.

Multiply $6$ by $0$ .

$f'' (0) = 0 + 6$

Add $0$ and $6$ .

$f'' (0) = 6$

The final answer is $6$ .

$6$

The graph is concave up on the interval $(- 1, \infty)$ because $f'' (0)$ is positive.

Concave up on $(- 1, \infty)$ since $f'' (x)$ is positive

Step 6

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave down on $(- \infty, - 1)$ since $f'' (x)$ is negative

Concave up on $(- 1, \infty)$ since $f'' (x)$ is positive

Step 7

Find the Concavity f(x)=x^3+3x^2-9x+5

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