Question

Assinale a alternativa cuja função é ímpar:

Se possível explicar como chegou ao resultado, eu tentei resolver e as questões A,D,E deu como impar na minha conta.

Answer 1

$\large\boxed{\begin{array}{l}\underline{\rm De\!\!~finic_{\!\!,}\tilde ao\,de\,func_{\!\!,}\tilde ao\,par}\\\sf Uma\,func_{\!\!,}\tilde ao\,\acute e\,dita\,par\,quando\,ocorre\\\sf f(-x)=f(x)\\\sf ou\,seja,a\,imagem\,oposta\,de\,x\,coincide\\\sf com\,a\,imagem\,de\,x.\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm De~\!\!finic_{\!\!,}\tilde ao \,de\,func_{\!\!,}\tilde ao\,\acute impar}\\\sf Uma\,func_{\!\!,}\tilde ao\,\acute e\,dita\,\acute impar\,quando\,ocorre\\\sf f(-x)=-f(x)\\\sf Ou\,seja,a\,imagem\,oposta\,de\,x\,\acute e\,o\,oposto\\\sf da\,imagem\,de\,x.\end{array}}$

$\large\boxed{\begin{array}{l}\sf Vamos\,verificar\,uma\,a\,uma\,para\,responder:\end{array}}$

$\large\boxed{\begin{array}{l}\rm a)~\sf f(x)=3x^2\\\sf f(-x)=3\cdot(-x)^2=3x^2=f(x)\\\implies\sf f(-x)=f(x)\longrightarrow func_{\!\!,}\tilde ao\,\acute e\,par\\\rm b)~\sf f(x)=x^4+x^2\\\sf f(-x)=(-x)^4+(-x)^2=x^4+x^2=f(x)\\\implies \sf f(-x)=f(x)\longrightarrow func_{\!\!,}\tilde ao\,\acute e\,par\end{array}}$

$\large\boxed{\begin{array}{l}\rm c)~\sf f(x)=5x^6\\\sf f(-x)=5\cdot(-x)^6=5x^6=f(x)\\\implies\sf f (-x)=f(x)\longrightarrow func_{\!\!,}\tilde ao\,\acute e\,par\\\rm d)~\sf f(x)=x^3-2x\\\sf f(-x)=(-x)^3-2(-x)=-x^3+2x\\\sf f(-x)=-(x^2-2x)=-f(x)\\\implies\sf f(-x)=-f(x)\longrightarrow func_{\!\!,}\tilde ao\,\acute impar\end{array}}$

$\large\boxed{\begin{array}{l}\rm e)~f(x)=(x^3-x)^2=x^6-2x^4+x^2\\\sf f(-x)=(-x)^6-2(-x)^4+(-x)^2\\\sf f(-x)=x^6-2x^4+x^2=f(x)\\\sf f(-x)=f(x)\longrightarrow func_{\!\!,}\tilde ao\,\acute e\,par\end{array}}$

$\Large\boxed{\begin{array}{l}\sf portanto\,o\,gabarito\,correto\,\acute e\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf\maltese\,alternativa\,d}}}}}\end{array}}$