Find the Function f(x)=3x^2-4x

$f (x) = 3 x^{2} - 4 x$

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

Split the single integral into multiple integrals.

$\int 3 x^{2} d x + \int - 4 x d x$

Step 3

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$3 \int x^{2} d x + \int - 4 x d x$

Step 4

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

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

Step 5

Since $- 4$ is constant with respect to $x$ , move $- 4$ out of the integral.

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

Step 6

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$3 (\frac{1}{3} x^{3} + C) - 4 (\frac{1}{2} x^{2} + C)$

Step 7

Simplify.

$3 (\frac{1}{3}) x^{3} - 4 (\frac{1}{2}) x^{2} + C$

Simplify.

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

$\frac{3}{3} x^{3} - 4 (\frac{1}{2}) x^{2} + C$

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3}{3} x^{3} - 4 (\frac{1}{2}) x^{2} + C$

Rewrite the expression.

$1 x^{3} - 4 (\frac{1}{2}) x^{2} + C$

Multiply $x^{3}$ by $1$ .

$x^{3} - 4 (\frac{1}{2}) x^{2} + C$

Combine $- 4$ and $\frac{1}{2}$ .

$x^{3} + \frac{- 4}{2} x^{2} + C$

Cancel the common factor of $- 4$ and $2$ .

Factor $2$ out of $- 4$ .

$x^{3} + \frac{2 \cdot - 2}{2} x^{2} + C$

Cancel the common factors.

Factor $2$ out of $2$ .

$x^{3} + \frac{2 \cdot - 2}{2 (1)} x^{2} + C$

Cancel the common factor.

$x^{3} + \frac{2 \cdot - 2}{2 \cdot 1} x^{2} + C$

Rewrite the expression.

$x^{3} + \frac{- 2}{1} x^{2} + C$

Divide $- 2$ by $1$ .

$x^{3} - 2 x^{2} + C$

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.

$F (x) = x^{3} - 2 x^{2} + C$

Find the Function f(x)=3x^2-4x

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