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]
