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Help me do this …math

Help Me Do This Math class=

Sagot :

Step-by-step explanation:

How to find inverse function:

Step 1: Write down equation.

[tex]f(x) = 4 \sin(x) + 3[/tex]

Remeber Ruler notation that

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

[tex]y = 4 \sin(x) + 3[/tex]

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

[tex]y - 3 = 4 \sin(x) [/tex]

Divide both sides by 4.

[tex] \frac{y - 3}{4} = \sin(x) [/tex]

[tex] \sin {}^{ - 1} ( \frac{y - 3}{4} ) = \sin {}^{ - 1} ( \sin(x) ) [/tex]

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

[tex] \sin {}^{ - 1} ( \frac{y - 3}{4} ) = x[/tex]

Swap x and y. So our inverse function is

[tex] \sin {}^{ - 1} ( \frac{x - 3}{4} ) = y[/tex]

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

[tex]x = \frac{\pi}{2} [/tex]

We then would get

[tex]4 \sin( \frac{\pi}{2 } ) + 3 = 4 \times (1) + 3 = 4 + 3 = 7[/tex]

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.

[tex] \sin {}^{ - 1} ( \frac{7 - 3}{4} ) = \sin {}^{ - 1} ( \frac{4}{4} ) = \sin {}^{ - 1} (1) = \frac{\pi}{2} [/tex]

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

[tex]f(x) = 4 \sin(x) + 3[/tex]

Parent function is

[tex] \sin(x) [/tex]

with a range of [-1,1].

We then vertical stretch by 4 so we get

[tex]4 \sin(x) [/tex]

and our range will be

[-4,4].

Then we add a vertical shift of 3.

[tex]4 \sin(x) + 3[/tex]

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]

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