Calculus Examples

y = 2 \sqrt{x} - x

$y = 2 \sqrt{x} - x$

Step 1

Reorder

2 \sqrt{x}

$2 \sqrt{x}$ and

- x

$- x$ .

y = - x + 2 \sqrt{x}

$y = - x + 2 \sqrt{x}$

Step 2

Find the domain for

y = 2 \sqrt{x} - x

$y = 2 \sqrt{x} - x$ so that a list of

x

$x$ values can be picked to find a list of points, which will help graphing the radical.

Tap for more steps...

Step 2.1

Set the radicand in

\sqrt{x}

$\sqrt{x}$ greater than or equal to

0

$0$ to find where the expression is defined.

x \geq 0

$x \geq 0$

Step 2.2

The domain is all values of

x

$x$ that make the expression defined.

Interval Notation:

[0, \infty)

$[0, \infty)$

Set-Builder Notation:

{x | x \geq 0}

${x | x \geq 0}$

Interval Notation:

[0, \infty)

$[0, \infty)$

Set-Builder Notation:

{x | x \geq 0}

${x | x \geq 0}$

Step 3

To find the radical expression end point, substitute the

x

$x$ value

0

$0$ , which is the least value in the domain, into

f (x) = - x + 2 \sqrt{x}

$f (x) = - x + 2 \sqrt{x}$ .

Tap for more steps...

Step 3.1

Replace the variable

x

$x$ with

0

$0$ in the expression.

f (0) = - (0) + 2 \sqrt{0}

$f (0) = - (0) + 2 \sqrt{0}$

Step 3.2

Simplify the result.

Tap for more steps...

Step 3.2.1

Simplify each term.

Tap for more steps...

Step 3.2.1.1

Multiply

- 1

$- 1$ by

0

$0$ .

f (0) = 0 + 2 \sqrt{0}

$f (0) = 0 + 2 \sqrt{0}$

Step 3.2.1.2

Rewrite

0

$0$ as

0^{2}

$0^{2}$ .

f (0) = 0 + 2 \sqrt{0^{2}}

$f (0) = 0 + 2 \sqrt{0^{2}}$

Step 3.2.1.3

Pull terms out from under the radical, assuming positive real numbers.

f (0) = 0 + 2 \cdot 0

$f (0) = 0 + 2 \cdot 0$

Step 3.2.1.4

Multiply

2

$2$ by

0

$0$ .

f (0) = 0 + 0

$f (0) = 0 + 0$

f (0) = 0 + 0

$f (0) = 0 + 0$

Step 3.2.2

Add

0

$0$ and

0

$0$ .

f (0) = 0

$f (0) = 0$

Step 3.2.3

The final answer is

0

$0$ .

0

$0$

0

$0$

0

$0$

Step 4

The radical expression end point is

(0, 0)

$(0, 0)$ .

(0, 0)

$(0, 0)$

Step 5

Select a few

x

$x$ values from the domain. It would be more useful to select the values so that they are next to the

x

$x$ value of the radical expression end point.

Tap for more steps...

Step 5.1

Substitute the

x

$x$ value

1

$1$ into

f (x) = - x + 2 \sqrt{x}

$f (x) = - x + 2 \sqrt{x}$ . In this case, the point is

(1, 1)

$(1, 1)$ .

Tap for more steps...

Step 5.1.1

Replace the variable

x

$x$ with

1

$1$ in the expression.

f (1) = - (1) + 2 \sqrt{1}

$f (1) = - (1) + 2 \sqrt{1}$

Step 5.1.2

Simplify the result.

Tap for more steps...

Step 5.1.2.1

Simplify each term.

Tap for more steps...

Step 5.1.2.1.1

Multiply

- 1

$- 1$ by

1

$1$ .

f (1) = - 1 + 2 \sqrt{1}

$f (1) = - 1 + 2 \sqrt{1}$

Step 5.1.2.1.2

Any root of

1

$1$ is

1

$1$ .

f (1) = - 1 + 2 \cdot 1

$f (1) = - 1 + 2 \cdot 1$

Step 5.1.2.1.3

Multiply

2

$2$ by

1

$1$ .

f (1) = - 1 + 2

$f (1) = - 1 + 2$

f (1) = - 1 + 2

$f (1) = - 1 + 2$

Step 5.1.2.2

Add

- 1

$- 1$ and

2

$2$ .

f (1) = 1

$f (1) = 1$

Step 5.1.2.3

The final answer is

1

$1$ .

y = 1

$y = 1$

y = 1

$y = 1$

y = 1

$y = 1$

Step 5.2

Substitute the

x

$x$ value

2

$2$ into

f (x) = - x + 2 \sqrt{x}

$f (x) = - x + 2 \sqrt{x}$ . In this case, the point is

(2, - 2 + 2 \sqrt{2})

$(2, - 2 + 2 \sqrt{2})$ .

Tap for more steps...

Step 5.2.1

Replace the variable

x

$x$ with

2

$2$ in the expression.

f (2) = - (2) + 2 \sqrt{2}

$f (2) = - (2) + 2 \sqrt{2}$

Step 5.2.2

Simplify the result.

Tap for more steps...

Step 5.2.2.1

Multiply

- 1

$- 1$ by

2

$2$ .

f (2) = - 2 + 2 \sqrt{2}

$f (2) = - 2 + 2 \sqrt{2}$

Step 5.2.2.2

The final answer is

- 2 + 2 \sqrt{2}

$- 2 + 2 \sqrt{2}$ .

y = - 2 + 2 \sqrt{2}

$y = - 2 + 2 \sqrt{2}$

y = - 2 + 2 \sqrt{2}

$y = - 2 + 2 \sqrt{2}$

y = - 2 + 2 \sqrt{2}

$y = - 2 + 2 \sqrt{2}$

Step 5.3

The square root can be graphed using the points around the vertex

(0, 0), (1, 1), (2, 0.83)

$(0, 0), (1, 1), (2, 0.83)$

\begin{array}{c} x & y \\ 0 & 0 \\ 1 & 1 \\ 2 & 0.83 \end{array}

$\begin{array}{c} x & y \\ 0 & 0 \\ 1 & 1 \\ 2 & 0.83 \end{array}$

\begin{array}{c} x & y \\ 0 & 0 \\ 1 & 1 \\ 2 & 0.83 \end{array}

$\begin{array}{c} x & y \\ 0 & 0 \\ 1 & 1 \\ 2 & 0.83 \end{array}$

Step 6