Problems from Symmetry, periodicity and intersection points of a function

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 $y=0\Rightarrow 0=f(x)=x^2-4\Rightarrow x^2=4\Rightarrow x=\pm 2\Rightarrow (2,0),(-2,0)$

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 $$f(-x)={\displaystyle \frac{-2x}{(-x{)}^2-1}}={\displaystyle \frac{-2x}{x^{-}1}}=-f(x)$$ 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 $y=0\Rightarrow 0=f(x)={\displaystyle \frac{2x}{x^2-1}}\Rightarrow 0=2x\Rightarrow x=0\Rightarrow (0,0)$

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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