Find the Function f(x)=(-x)^(1/2)

$f (x) = {(- x)}^{\frac{1}{2}}$

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

Simplify.

Factor $- 1$ out of $- x$ .

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

Apply the product rule to $- (x)$ .

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

Step 3

Since ${(- 1)}^{\frac{1}{2}}$ is constant with respect to $x$ , move ${(- 1)}^{\frac{1}{2}}$ out of the integral.

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

Step 4

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

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

Step 5

Simplify the answer.

Rewrite ${(- 1)}^{\frac{1}{2}} (\frac{2}{3} x^{\frac{3}{2}} + C)$ as $i (\frac{2}{3} x^{\frac{3}{2}}) + C$ .

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

Simplify.

Combine $\frac{2}{3}$ and $x^{\frac{3}{2}}$ .

$i \frac{2 x^{\frac{3}{2}}}{3} + C$

Combine $i$ and $\frac{2 x^{\frac{3}{2}}}{3}$ .

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

Move $2$ to the left of $i$ .

$\frac{2 i x^{\frac{3}{2}}}{3} + C$

Reorder terms.

$\frac{2 i}{3} x^{\frac{3}{2}} + C$

Step 6

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

$F (x) = \frac{2 i}{3} x^{\frac{3}{2}} + C$