Find the Function f(x)=9-6x

$f (x) = 9 - 6 x$

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.

$\int 9 d x + \int - 6 x d x$

Step 3

Apply the constant rule.

$9 x + C + \int - 6 x d x$

Step 4

Since $- 6$ is constant with respect to $x$ , move $- 6$ out of the integral.

$9 x + C - 6 \int x d x$

Step 5

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

$9 x + C - 6 (\frac{1}{2} x^{2} + C)$

Step 6

Simplify.

$9 x - 6 (\frac{1}{2}) x^{2} + C$

Simplify.

Combine $- 6$ and $\frac{1}{2}$ .

$9 x + \frac{- 6}{2} x^{2} + C$

Cancel the common factor of $- 6$ and $2$ .

Factor $2$ out of $- 6$ .

$9 x + \frac{2 \cdot - 3}{2} x^{2} + C$

Cancel the common factors.

Factor $2$ out of $2$ .

$9 x + \frac{2 \cdot - 3}{2 (1)} x^{2} + C$

Cancel the common factor.

$9 x + \frac{2 \cdot - 3}{2 \cdot 1} x^{2} + C$

Rewrite the expression.

$9 x + \frac{- 3}{1} x^{2} + C$

Divide $- 3$ by $1$ .

$9 x - 3 x^{2} + C$

Step 7

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

$F (x) = 9 x - 3 x^{2} + C$