Say if the following functions are symmetric, antisymmetric and/or periodic and find the intersection points of the functions with the axes:
- $$f(x)=x^2-4$$
- $$f(x)=\cos (x)$$
- $$f(x)=\dfrac{2x}{x^2-1}$$
- $$f(x)=x$$
Development:
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The function is symmetric with respect to the axis $$x=0$$: $$$ f(-x)=(-x)^2-4=x^2-4=f(x)$$$ The function is not periodic since it does not repeat itself.
Intersection points with the axes:
Si $$x=0 \Rightarrow y=f(0)=-4 \Rightarrow (0,-4)$$
Si $$y=0 \Rightarrow 0=f(x)=x^2-4 \Rightarrow x^2=4 \Rightarrow x=\pm2 \Rightarrow (2,0), \ (-2,0)$$
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The function is symmetric with respect to the axis $$x=0$$ since $$$\cos (x)=\cos(-x)$$$ Also, the cosine is $$2\pi$$-periodic: $$\cos(x+2\pi)=\cos(x)$$.
Intersection points with the axes:
If $$x=0 \Rightarrow y=f(0)=\cos(0)=1 \Rightarrow (0,1)$$
If $$y=0 \Rightarrow 0=f(x)=\cos(x) \Rightarrow x=\pi+\pi k \Rightarrow (\pi+\pi k,0)$$ for $$k\in\mathbb{Z}$$
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This function is antisymmetric with respect to the axis $$x=0$$ since $$$ f(-x)=\dfrac{-2x}{(-x)^2-1}=\dfrac{-2x}{x^-1}=-f(x)$$$ It does not present any type of period.
Intersection points with the axes:
If $$x=0 \Rightarrow y=f(0)=0 \Rightarrow (0,0)$$
If $$y=0 \Rightarrow 0=f(x)=\dfrac{2x}{x^2-1} \Rightarrow 0=2x \Rightarrow x=0 \Rightarrow (0,0)$$
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Clearly antisymmetrical function in the axis $$x=0$$: $$$f(-x)=-x=-f(x)$$$ Intersection points with the axes:
If $$x=0 \Rightarrow y=f(0)=0 \Rightarrow (0,0)$$
If $$y=0 \Rightarrow 0=f(x)=x \Rightarrow x=0 \Rightarrow (0,0)$$
Solution:
- Even function, not periodic. Intersects with the axes at points $$(0,-4),\ (2,0), \ (-2,0)$$.
- Even function, periodic of period $$2\pi$$. Intersects with the axes at points $$(0,1), \ (\pi+\pi k,0)$$ for $$k\in\mathbb{Z}$$.
- Odd function, without periods. Intersects with the axes at points $$(0,0)$$.
- Odd function, without periods. Intersects with the axes at points $$(0,0)$$.