Question

Which of the following inverse functions are defined for x = - 1/2? Select 4 of the following that apply. There must be 4 selections!

Answer 1

Answer:

$y=cos^{-1}(x)$,

$y=cot^{-1}(-\frac{1}{2})$,

$y=sin^{-1}(-\frac{1}{2})$

$y=tan^{-1}(-\frac{1}{2})$

Above four functions are defined at the given point.

Step-by-step explanation:

We have been given all trigonometric function we need to tell which among them is defined for $x=-\frac{1}{2}$

Case 1: $y=cos^{-1}(x)$

Since, $At x=-\frac{1}{2}$

$y=cos^{-1}(-\frac{1}{2})$

$y=-cos^{-1}(\frac{1}{2})$

$y=\frac{2\pi}{3}$

Domain of the function is $[-1,1]$

Defined at given point.

Case 2: $y=cot^{-1}(-\frac{1}{2})$

$y=-cot^{-1}(\frac{1}{2})$

$y=-1.1071$

Domain of the function is $(-\infty,\infty)$

Defined at given point.

Case 3: $y=cosec^{-1}(-\frac{1}{2})$

$y=-cosec^{-1}(\frac{1}{2})$

Value of the function is not defined because it is out of the domain of the function.

Domain is $(-\infty,-1]\cup[1,\infty)$

So, not defined at given point.

Case 4: $y=sec^{-1}(-\frac{1}{2})$

$y=-sec^{-1}(\frac{1}{2})$

Value of the function is not defined because it is out of the domain of the function.

Domain is $(-\infty,-1]\cup[1,\infty)$

So, not defined at given point.

Case 5:$y=sin^{-1}(-\frac{1}{2})$

$y=-sin^{-1}(\frac{1}{2})$

$y=\frac{-\pi}{6}$

Domain of the function is $[-1,1]$

Defined at given point.

Case 6:$y=tan^{-1}(-\frac{1}{2})$

$y=-tan^{-1}(\frac{1}{2})$

$y=-0.4636$

Domain of the function is $(-\infty,\infty)$

Defined at given point.

Therefore, Option 1,2,5,6 are correct.

Answer 2

Answer: I just took the test and got 100%

1. B, f(x) = 4 sin x/2 -3

2. C = 1; A = 2; B = 2

3. C, H(t) = -2.4 cos (0.017t) +12

4. A, y = cos ^ -1 x; B, y = cot ^ -1 x; E, y = sin ^ -1 x; F, y = tan ^ -1 x

5. C, y = sin ^1 x

6. C, 48.7*

7. C, f(g(x)) = sec(sin x)

    Domain: all real numbers

    Range: 1 < x < 1.85

8. C, step 3

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