a)

Obliczmy w jakich punktach wykres przecina oś OX.

$x^4-4x^2=0$  

$x^2\left(x^2-4\right)=0$  

$x=0\vee x=2\vee x=-2$  

 

$A=\left(0,0\right)$  

$B=\left(2,0\right)$  

$C=\left(-2,0\right)$  

 

$f{}^{\prime }\left(x\right)=4x^3-8x$  

 

Obliczenia dla punktu A:

$f\left(0\right)=0$  

$f{}^{\prime }\left(0\right)=0$  

$y-f\left(x_0\right)=f{}^{\prime }\left(x_0\right)\left(x-x_0\right)$  

$y-0=0\left(x-0\right)$  

$y=0$  

 

Obliczenia dla punktu B:

$f\left(2\right)=2^4-4\cdot 2^2=16-4\cdot 4=0$  

$f{}^{\prime }\left(2\right)=4\cdot 2^3-8\cdot 2=4\cdot 8-16=32-16=16$  

$y-f\left(x_0\right)=f{}^{\prime }\left(x_0\right)\left(x-x_0\right)$  

$y-0=16\left(x-2\right)$  

$y=16x-32$  

 

Obliczenia dla punktu C:

$f\left(-2\right)={\left(-2\right)}^4-4\cdot {\left(-2\right)}^2=16-4\cdot 4=0$  

$f{}^{\prime }\left(-2\right)=4\cdot {\left(-2\right)}^3-8\cdot \left(-2\right)=4\cdot \left(-8\right)+16=-32+16=-16$  

$y-f\left(x_0\right)=f{}^{\prime }\left(x_0\right)\left(x-x_0\right)$  

$y-0=-16\left(x-\left(-2\right)\right)$  

$y=-16\left(x+2\right)$  

$y=-16x-32$  

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b)

Obliczmy w jakich punktach wykres przecina oś OX.

$\frac{x^2-1}{x^2+1}=0$  

$x^2-1=0$  

$\left(x-1\right)\left(x+1\right)=0$  

$x=1\vee x=-1$   

 

$A=\left(1,0\right)$  

$B=\left(-1,0\right)$  

 

$f{}^{\prime }\left(x\right)=\frac{\left(x^2-1\right){}^{\prime }\left(x^2+1\right)-\left(x^2-1\right)\left(x^2+1\right){}^{\prime }}{{\left(x^2+1\right)}^2}=\frac{2x\left(x^2+1\right)-\left(x^2-1\right)\cdot 2x}{{\left(x^2+1\right)}^2}=\frac{2x^3+2x-2x^3+2x}{{\left(x^2+1\right)}^2}=\frac{4x}{{\left(x^2+1\right)}^2}$  

 

Obliczenia dla punktu A:

$f\left(1\right)=\frac{1^2-1}{1^2+1}=0$  

$f{}^{\prime }\left(1\right)=\frac{4\cdot 1}{{\left(1^2+1\right)}^2}=\frac{4}{4}=1$  

$y-f\left(x_0\right)=f{}^{\prime }\left(x_0\right)\left(x-x_0\right)$  

$y-0=1\left(x-1\right)$  

$y=x-1$  

 

Obliczenia dla punktu B:

$f\left(-1\right)=\frac{{\left(-1\right)}^2-1}{{\left(-1\right)}^2+1}=0$

   $f{}^{\prime }\left(-1\right)=\frac{4\cdot \left(-1\right)}{{\left({\left(-1\right)}^2+1\right)}^2}=\frac{-4}{4}=-1$   

$y-f\left(x_0\right)=f{}^{\prime }\left(x_0\right)\left(x-x_0\right)$  

$y-0=-1\left(x-\left(-1\right)\right)$  

$y=-1\left(x+1\right)$  

$y=-x-1$