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Sagot :
Well the period of the tan(x) function is π. It means that when you add π on X axis you will get the same y axis value.
There is a little trick to find the period of functions like yours.
You just divide period of basic tan(x) function with, in your case 3.
So this mean you have
To = π/3
Function intercepts x axis when function is equal to 0.
tan(3x) = 0
3x = k * π
x = k * (π / 3)
There are only vertical asymptotes in tangent function.
You can easily see it if you look at the graph
tan(x) = ∞ , (π/2 + kπ)
tan(x) = -∞ (-π/2 + kπ).
x = π/2 + kπ
x = - π/2 + kπ
tan(3x) = ∞
3x = π/2 + kπ
x = π/6 + kπ/3
tan(3x) = -∞
3x = -π/2 + kπ
x = -π/6 + kπ/3
There is a little trick to find the period of functions like yours.
You just divide period of basic tan(x) function with, in your case 3.
So this mean you have
To = π/3
Function intercepts x axis when function is equal to 0.
tan(3x) = 0
3x = k * π
x = k * (π / 3)
There are only vertical asymptotes in tangent function.
You can easily see it if you look at the graph
tan(x) = ∞ , (π/2 + kπ)
tan(x) = -∞ (-π/2 + kπ).
x = π/2 + kπ
x = - π/2 + kπ
tan(3x) = ∞
3x = π/2 + kπ
x = π/6 + kπ/3
tan(3x) = -∞
3x = -π/2 + kπ
x = -π/6 + kπ/3
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