Trigonometry Examples

$f (x) = \sqrt{\ln (y - 2 x)} + \arcsin (x + 1)$

Step 1

The domain in terms of $y$ are all the $y$ values that make the radicand non-negative.

$[- 2, - 1]$

${y | - 2 \leq y \leq - 1}$

Step 2

To find the end points, substitute the bounds of the $y$ values from the domain into $f (y) = \sqrt{\ln (y - 2 y)} + \arcsin (y + 1)$ .

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Step 2.1

Replace the variable $y$ with $S E T, x | - 2$ in the expression.

$f (S E T, x | - 2) = \sqrt{\ln ((S E T, x | - 2) - 2 (S E T, x | - 2))} + \arcsin ((S E T, x | - 2) + 1)$

Step 2.2

Simplify the result.

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Step 2.2.1

Multiply $- 2$ by each element of the matrix.

$f (S E T, x | - 2) = \sqrt{\ln (((S E T, x | - 2) - 2 S E T, x | - 2))} + \arcsin ((S E T, x | - 2) + 1)$

Step 2.2.2

The final answer is $\sqrt{\ln (((S E T, x | - 2) - 2 S E T, x | - 2))} + \arcsin ((S E T, x | - 2) + 1)$ .

$\sqrt{\ln (((S E T, x | - 2) - 2 S E T, x | - 2))} + \arcsin ((S E T, x | - 2) + 1)$

Step 2.3

Replace the variable $y$ with $- 1$ in the expression.

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

Step 2.4

Simplify the result.

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Step 2.4.1

Simplify each term.

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Step 2.4.1.1

Multiply $- 2$ by $- 1$ .

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

Step 2.4.1.2

Add $- 1$ and $2$ .

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

Step 2.4.1.3

The natural logarithm of $1$ is $0$ .

$f (- 1) = \sqrt{0} + \arcsin ((- 1) + 1)$

Step 2.4.1.4

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

$f (- 1) = \sqrt{0^{2}} + \arcsin ((- 1) + 1)$

Step 2.4.1.5

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

$f (- 1) = 0 + \arcsin ((- 1) + 1)$

Step 2.4.1.6

Add $- 1$ and $1$ .

$f (- 1) = 0 + \arcsin (0)$

Step 2.4.1.7

The exact value of $\arcsin (0)$ is $0$ .

$f (- 1) = 0 + 0$

Step 2.4.2

Add $0$ and $0$ .

$f (- 1) = 0$

Step 2.4.3

The final answer is $0$ .

$0$

Step 3

The end points are $(\sqrt{\ln (((S E T, x | - 2) - 2 S E T, x | - 2))} + \arcsin ((S E T, x | - 2) + 1), S E T, x | - 2), (0, - 1)$ .

$(\sqrt{\ln (((S E T, x | - 2) - 2 S E T, x | - 2))} + \arcsin ((S E T, x | - 2) + 1), S E T, x | - 2), (0, - 1)$

Step 4

The square root can be graphed using the points around the vertex $(\sqrt{\ln (((S E T, x | - 2) - 2 S E T, x | - 2))} + \arcsin ((S E T, x | - 2) + 1), S E T, x | - 2), (0, - 1),$

Step 5