What is the inverse of f(x)=9sin(x-5)+8 ?

The inverse of f(x)=9sin(x-5)+8 is arcsin((x-8)/9)+5

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inverse of f(x)=9sin(x-5)+8

inverse

f (x) = 9 sin (x - 5) + 8

arcsin (\frac{x - 8}{9}) + 5

Solution steps

Solve for

sin (x - 5), y = 9 sin (x - 5) + 8

sin (x - 5) = \frac{y - 8}{9}

sin (a) = b \Rightarrow a = arcsin (b)

x - 5 = arcsin (\frac{y - 8}{9})

Solve for

x, x - 5 = arcsin (\frac{y - 8}{9})

arcsin (\frac{y - 8}{9}) + 5

Substitute

y = x

arcsin (\frac{x - 8}{9}) + 5

f (x) = \frac{2 x - 6}{4}

f (x) = x^{2} - 12 x + 11

f (x) = \frac{2 x + 1}{4 x - 1}

f (x) = 3 x^{2} - 2 x + 1

f (x) = (x + 1)^{2} - 2, x \geq - 1

What is the inverse of f(x)=9sin(x-5)+8 ?
The inverse of f(x)=9sin(x-5)+8 is arcsin((x-8)/9)+5