Find the Function f(x)=|-2x-3|

$f (x) = | - 2 x - 3 |$

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

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

Set the argument in the absolute value equal to $0$ to find the potential values to split the solution at.

$- 2 x - 3 = 0$

Solve the equation for $x$ .

$x = - \frac{3}{2}$

Create intervals around the solutions to find where $- 2 x - 3$ is positive and negative.

$(- \infty, - \frac{3}{2}), (- \frac{3}{2}, \infty)$

Substitute a value from each interval into $- 2 x - 3$ to figure out where the expression is positive or negative.

$\begin{array}{c} Interval & Sign on interval \\ (- \infty, - \frac{3}{2}) & + \\ (- \frac{3}{2}, \infty) & - \end{array}$

Integrate the argument of the absolute value.

Set up the integral with the argument of the absolute value.

$\int - 2 x - 3 d x$

Split the single integral into multiple integrals.

$\int - 2 x d x + \int - 3 d x$

Since $- 2$ is constant with respect to $x$ , move $- 2$ out of the integral.

$- 2 \int x d x + \int - 3 d x$

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

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

Apply the constant rule.

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

Combine $\frac{1}{2}$ and $x^{2}$ .

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

Simplify.

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

On the intervals where the argument is negative, multiply the solution of the integral by $- 1$ .

${\begin{cases} - x^{2} - 3 x + C & x \leq - \frac{3}{2} \\ - (- x^{2} - 3 x + C) & x > - \frac{3}{2} \end{cases}$

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

$F (x) = {\begin{cases} - x^{2} - 3 x + C & x \leq - \frac{3}{2} \\ - (- x^{2} - 3 x + C) & x > - \frac{3}{2} \end{cases}$