[Derivada] Achar pontos de inflexão

por **alienpuke** » Qui Nov 12, 2015 11:31

Olá, gostaria de saber se essa segunda derivada possui algum ponto de inflexão e se não houver o porquê. Obrigado!

$f(x)=\frac{{6x}^{2}+2}{({1-{x}^{2}})^{3}}$

Ps. Tentei igualar a 0 mas nao achei raízes reais. Por esse motivo eu não tenho pontos de inflexão?

por **Baltuilhe** » Sáb Nov 14, 2015 18:14

Boa tarde!!

Calculando a derivada primeira:
$f(x)=\frac{6x^2+2}{\left(1-x^2\right)^3}\\ f'(x)=\frac{\left(6x^2+2\right)'\cdot\left(1-x^2\right)^3-\left(6x^2+2\right)\cdot\left[\left(1-x^2\right)^3\right]'}{\left[\left(1-x^2\right)^3\right]^2}\\ f'(x)=\frac{\left(12x\right)\cdot\left(1-x^2\right)^3-\left(6x^2+2\right)\cdot\left[3\left(1-x^2\right)^2\cdot\left(1-x^2\right)'\right]}{\left(1-x^2\right)^6}\\ f'(x)=\frac{\left(1-x^2\right)^2\left[\left(12x\right)\cdot\left(1-x^2\right)-\left(6x^2+2\right)\cdot 3\cdot\left(-2x\right)\right]}{\left(1-x^2\right)^6}\\ f'(x)=\frac{12x-12x^3+36x^3+12x}{\left(1-x^2\right)^4}\\ f'(x)=\frac{24x^3+24x}{\left(1-x^2\right)^4}\\ f'(x)=\frac{24x\cdot\left(x^2+1\right)}{\left(1-x^2\right)^4}\\$

Agora podemos calcular a derivada segunda:
$f'(x)=\frac{24x\cdot\left(x^2+1\right)}{\left(1-x^2\right)^4}\\ f''(x)=\frac{\left[24x\cdot\left(x^2+1\right)\right]'\cdot\left(1-x^2\right)^4-24x\cdot\left(x^2+1\right)\cdot\left[\left(1-x^2\right)^4\right]'}{\left[\left(1-x^2\right)^4\right]^2}\\ f''(x)=\frac{\left(72x^2+24\right)\cdot\left(1-x^2\right)^4-24x\cdot\left(x^2+1\right)\cdot\left[4\left(1-x^2\right)^3\cdot\left(1-x^2\right)'\right]}{\left(1-x^2\right)^8}\\ f''(x)=\frac{24\left(3x^2+1\right)\cdot\left(1-x^2\right)^4-24x\cdot\left(x^2+1\right)\cdot\left[4\left(1-x^2\right)^3\cdot\left(-2x\right)\right]}{\left(1-x^2\right)^8}\\ f''(x)=\frac{\left(1-x^2\right)^3\cdot\left[24\left(3x^2+1\right)\cdot\left(1-x^2\right)-24x\cdot\left(x^2+1\right)\cdot 4\cdot\left(-2x\right)\right]}{\left(1-x^2\right)^8}\\ f''(x)=\frac{24\left[\left(3x^2+1\right)\cdot\left(1-x^2\right)+8x^2\cdot\left(x^2+1\right)\right]}{\left(1-x^2\right)^5}\\ f''(x)=\frac{24\left(3x^2-3x^4+1-x^2+8x^4+8x^2\right)}{\left(1-x^2\right)^5}$
$f''(x)=\frac{24\left(5x^4+10x^2+1\right)}{\left(1-x^2\right)^5}\\$

De posse das derivadas consegue resolver o problema, certo?
Calcule as raízes da equação bi-quadrada.
Espero ter ajudado!