What is the periodicity of f(x)=4sin(1/pi x-2)+8 ?

The periodicity of f(x)=4sin(1/pi x-2)+8 is 2pi^2

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periodicity of f(x)=4sin(1/pi x-2)+8

periodicity

f (x) = 4 sin (\frac{1}{π} x - 2) + 8

2 π^{2}

Solution steps

Periodicity of

a \cdot sin (b x + c) + d = \frac{periodicity of sin ( x )}{∣ b ∣}

Periodicity of

sin (x)

2 π

= \frac{2 π}{\frac{1}{π}}

Simplify

= 2 π^{2}

f (x) = 2 x^{2} + 5 x - 3

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

f (x) = \frac{- 5 x}{3 x + 5}

x - 4 y = 4

y = e^{x} + 2 e^{2 x}

What is the periodicity of f(x)=4sin(1/pi x-2)+8 ?
The periodicity of f(x)=4sin(1/pi x-2)+8 is 2pi^2