Find the Local Maxima and Minima f(x)=x^3+3x^2+6x-12

$f (x) = x^{3} + 3 x^{2} + 6 x - 12$

Step 1

Find the first derivative of the function.

Differentiate.

By the Sum Rule, the derivative of $x^{3} + 3 x^{2} + 6 x - 12$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 12]$ .

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 12]$

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

$3 x^{2} + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 12]$

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

$3 x^{2} + 3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 12]$

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

$3 x^{2} + 3 (2 x) + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 12]$

Multiply $2$ by $3$ .

$3 x^{2} + 6 x + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 12]$

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

$3 x^{2} + 6 x + 6 \frac{d}{d x} [x] + \frac{d}{d x} [- 12]$

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

$3 x^{2} + 6 x + 6 \cdot 1 + \frac{d}{d x} [- 12]$

Multiply $6$ by $1$ .

$3 x^{2} + 6 x + 6 + \frac{d}{d x} [- 12]$

Differentiate using the Constant Rule.

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12$ with respect to $x$ is $0$ .

$3 x^{2} + 6 x + 6 + 0$

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

$3 x^{2} + 6 x + 6$

Step 2

Find the second derivative of the function.

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

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

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

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

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

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

Multiply $2$ by $3$ .

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

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

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

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

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

Multiply $6$ by $1$ .

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

Differentiate using the Constant Rule.

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

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

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

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

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$3 x^{2} + 6 x + 6 = 0$

Step 4

Since there is no value of $x$ that makes the first derivative equal to $0$ , there are no local extrema.

No Local Extrema

Step 5

No Local Extrema

Step 6

Find the Local Maxima and Minima f(x)=x^3+3x^2+6x-12

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