Question

Help me do this …math

Answer 1

Step-by-step explanation:

How to find inverse function:

Step 1: Write down equation.

$f(x) = 4 \sin(x) + 3$

Remeber Ruler notation that

f(x)=y so we replace y with f(x).

$y = 4 \sin(x) + 3$

Next, we must isolate x so first we subtract 3.

$y - 3 = 4 \sin(x)$

Divide both sides by 4.

$\frac{y - 3}{4} = \sin(x)$

$\sin {}^{ - 1} ( \frac{y - 3}{4} ) = \sin {}^{ - 1} ( \sin(x) )$

Remeber that sin^-1x and sin x are inverse functions so they will cancel out to x. So we get

$\sin {}^{ - 1} ( \frac{y - 3}{4} ) = x$

Swap x and y. So our inverse function is

$\sin {}^{ - 1} ( \frac{x - 3}{4} ) = y$

If you want another proof: Here's one,

Let plug an a x value for the orginal equation,

4 sin x+3. Let say that

$x = \frac{\pi}{2}$

We then would get

$4 \sin( \frac{\pi}{2 } ) + 3 = 4 \times (1) + 3 = 4 + 3 = 7$

So when x=pi/2, y=7.

By definition of a inverse function, if we let 7 be our input, we should get pi/2. as a output.

So let see.

$\sin {}^{ - 1} ( \frac{7 - 3}{4} ) = \sin {}^{ - 1} ( \frac{4}{4} ) = \sin {}^{ - 1} (1) = \frac{\pi}{2}$

So this is the inverse function of 4 sin x+3.

1b. The range of f(x) is [-1,7).

We can use transformations to describe range.

We have

$f(x) = 4 \sin(x) + 3$

Parent function is

$\sin(x)$

with a range of [-1,1].

We then vertical stretch by 4 so we get

$4 \sin(x)$

and our range will be

[-4,4].

Then we add a vertical shift of 3.

$4 \sin(x) + 3$

So our range of 4 sin x+3.

[-1,7].

1c. Domain of a inverse function is the range of the orginal function.

The range of f(x) is [-1,7) so the domain of f^-1(x) is [-1,7].

f(x) domain was restricted to -pi/2 to pi/2 so the range of f^-1(x) is [-pi/2, pi/2]