Calculus Examples

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

Step 1

Reorder $2 \sqrt{x}$ and $- x$ .

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

Step 2

Find the domain for $y = 2 \sqrt{x} - x$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the radical.

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Step 2.1

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

Step 2.2

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Step 3

To find the radical expression end point, substitute the $x$ value $0$ , which is the least value in the domain, into $f (x) = - x + 2 \sqrt{x}$ .

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Step 3.1

Replace the variable $x$ with $0$ in the expression.

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

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Multiply $- 1$ by $0$ .

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

Step 3.2.1.2

Rewrite $0$ as $0^{2}$ .

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

Step 3.2.1.3

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

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

Step 3.2.1.4

Multiply $2$ by $0$ .

$f (0) = 0 + 0$

Step 3.2.2

Add $0$ and $0$ .

$f (0) = 0$

Step 3.2.3

The final answer is $0$ .

$0$

Step 4

The radical expression end point is $(0, 0)$ .

$(0, 0)$

Step 5

Select a few $x$ values from the domain. It would be more useful to select the values so that they are next to the $x$ value of the radical expression end point.

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Step 5.1

Substitute the $x$ value $1$ into $f (x) = - x + 2 \sqrt{x}$ . In this case, the point is $(1, 1)$ .

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Step 5.1.1

Replace the variable $x$ with $1$ in the expression.

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

Step 5.1.2

Simplify the result.

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Step 5.1.2.1

Simplify each term.

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Step 5.1.2.1.1

Multiply $- 1$ by $1$ .

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

Step 5.1.2.1.2

Any root of $1$ is $1$ .

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

Step 5.1.2.1.3

Multiply $2$ by $1$ .

$f (1) = - 1 + 2$

Step 5.1.2.2

Add $- 1$ and $2$ .

$f (1) = 1$

Step 5.1.2.3

The final answer is $1$ .

$y = 1$

Step 5.2

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

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Step 5.2.1

Replace the variable $x$ with $2$ in the expression.

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

Step 5.2.2

Simplify the result.

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Step 5.2.2.1

Multiply $- 1$ by $2$ .

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

Step 5.2.2.2

The final answer is $- 2 + 2 \sqrt{2}$ .

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

Step 5.3

The square root can be graphed using the points around the vertex $(0, 0), (1, 1), (2, 0.83)$

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

Step 6