Find the Function f(x)=(3x-1)^3

$f (x) = {(3 x - 1)}^{3}$

Step 1

The function $F (x)$ can be found by evaluating the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 2

Let $u = 3 x - 1$ . Then $d u = 3 d x$ , so $\frac{1}{3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

Let $u = 3 x - 1$ . Find $\frac{d u}{d x}$ .

Differentiate $3 x - 1$ .

$\frac{d}{d x} [3 x - 1]$

By the Sum Rule, the derivative of $3 x - 1$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [3 x] + \frac{d}{d x} [- 1]$

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

$3 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

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

$3 \cdot 1 + \frac{d}{d x} [- 1]$

Multiply $3$ by $1$ .

$3 + \frac{d}{d x} [- 1]$

Differentiate using the Constant Rule.

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$3 + 0$

Add $3$ and $0$ .

$3$

Rewrite the problem using $u$ and $d u$ .

$\int u^{3} \frac{1}{3} d u$

Step 3

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

$\int \frac{u^{3}}{3} d u$

Step 4

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{3} \int u^{3} d u$

Step 5

By the Power Rule, the integral of $u^{3}$ with respect to $u$ is $\frac{1}{4} u^{4}$ .

$\frac{1}{3} (\frac{1}{4} u^{4} + C)$

Step 6

Simplify.

Rewrite $\frac{1}{3} (\frac{1}{4} u^{4} + C)$ as $\frac{1}{3} \cdot \frac{1}{4} u^{4} + C$ .

$\frac{1}{3} \cdot \frac{1}{4} u^{4} + C$

Simplify.

Multiply $\frac{1}{3}$ by $\frac{1}{4}$ .

$\frac{1}{3 \cdot 4} u^{4} + C$

Multiply $3$ by $4$ .

$\frac{1}{12} u^{4} + C$

Step 7

Replace all occurrences of $u$ with $3 x - 1$ .

$\frac{1}{12} {(3 x - 1)}^{4} + C$

Step 8

The function $F$ if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

$F (x) = \frac{1}{12} \cdot {(3 x - 1)}^{4} + C$

Find the Function f(x)=(3x-1)^3

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