Question

Determine a expressão designatória da função f' (-1) para:
(segue em anexo)

ALGUÉM SABE FAZER ISSO? PELO AMOR DE DEUS TO ME MATANDO AQUI E NAO CONSIGO RESOLVER ISSO :(

Answer 1

Resolvemos os determinantes de matrizes de ordem 2 dessa forma:

$\left|\begin{array}{cc}a&d\\c&b\end{array}\right|=ab-cd$
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a)

$f(x)=\left|\begin{array}{cc}x^{3}&2\\3x+5&2x+1\end{array}\right|\\\\\\f(x)=x^{3}(2x+1)-2(3x+5)\\\\f(x)=2x^{4}+x^{3}-6x-10$

Derivando f(x):

$f'(x)=4\cdot2x^{4-1}+3x^{3-1}-1\cdot6x^{1-1}-0\\f'(x)=8x^{3}+3x^{2}-6$

Achando f'(-1):

$f'(-1)=8(-1)^{3}+3(-1)^{2}-6\\f'(-1)=8(-1)+3(1)-6\\f'(-1)=-8+3-6\\f'(-1)=-11$

b)

$f(x)=\left|\begin{array}{cc}\frac{2}{3}x+\frac{1}{2}&-7\\x-1&x^{2}\end{array}\right|\\\\\\f(x)=(\frac{2}{3}x+\frac{1}{2})x^{2}-(-7)(x-1)\\\\f(x)=\frac{2}{3}x^{3}+\frac{1}{2}x^{2}+7(x-1)\\\\f(x)=\frac{2}{3}x^{3}+\frac{1}{2}x^{2}+7x-7$

Derivando:

$f'(x)=3\cdot\frac{2}{3}x^{3-1}+2\cdot\frac{1}{2}x^{2-1}+1\cdot7x^{1-1}-0\\\\f'(x)=2x^{2}+x+7$

Achando f'(-1):

$f'(-1)=2(-1)^{2}+(-1)+7\\\\f'(-1)=2(1)-1+7\\\\f'(-1)=8$