# Problems from Symmetry, periodicity and intersection points of a function

Say if the following functions are symmetric, antisymmetric and/or periodic and find the intersection points of the functions with the axes:

1. $$f(x)=x^2-4$$
2. $$f(x)=\cos (x)$$
3. $$f(x)=\dfrac{2x}{x^2-1}$$
4. $$f(x)=x$$
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### Development:

1. 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 $$y=0 \Rightarrow 0=f(x)=x^2-4 \Rightarrow x^2=4 \Rightarrow x=\pm2 \Rightarrow (2,0), \ (-2,0)$$ 2. 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 $$k\in\mathbb{Z}$$

3. 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 $$y=0 \Rightarrow 0=f(x)=\dfrac{2x}{x^2-1} \Rightarrow 0=2x \Rightarrow x=0 \Rightarrow (0,0)$$ 4. 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:

1. Even function, not periodic. Intersects with the axes at points $$(0,-4),\ (2,0), \ (-2,0)$$.
2. Even function, periodic of period $$2\pi$$. Intersects with the axes at points $$(0,1), \ (\pi+\pi k,0)$$ for $$k\in\mathbb{Z}$$.
3. Odd function, without periods. Intersects with the axes at points $$(0,0)$$.
4. Odd function, without periods. Intersects with the axes at points $$(0,0)$$.
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