ressolução:f(x) = (x – 2)(x + 2) + (x + 1)(x – 1), ponto de abscissa x0 = 2. Intervalos: -4 ≤ x ≤ + 4 e -15 ≤ y≤ + 30
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Boa tarde!
Solução!
![f(x)=(x-2)\times(x+2)+(x+1)\times(x-1)\\\\\\\
f(x)=( x^{2} +2x-2x-4)+( x^{2} -x+x-1)\\\\\\
f(x)= x^{2} +0-4+ x^{2} +0-1\\\\\\
f(x)=2 x^{2} -5](https://tex.z-dn.net/?f=f%28x%29%3D%28x-2%29%5Ctimes%28x%2B2%29%2B%28x%2B1%29%5Ctimes%28x-1%29%5C%5C%5C%5C%5C%5C%5C%0Af%28x%29%3D%28+x%5E%7B2%7D+%2B2x-2x-4%29%2B%28+x%5E%7B2%7D+-x%2Bx-1%29%5C%5C%5C%5C%5C%5C%0Af%28x%29%3D+x%5E%7B2%7D+%2B0-4%2B+x%5E%7B2%7D+%2B0-1%5C%5C%5C%5C%5C%5C%0Af%28x%29%3D2+x%5E%7B2%7D+-5 "f(x)=(x-2)\times(x+2)+(x+1)\times(x-1)\\\\\\\
f(x)=( x^{2} +2x-2x-4)+( x^{2} -x+x-1)\\\\\\
f(x)= x^{2} +0-4+ x^{2} +0-1\\\\\\
f(x)=2 x^{2} -5")
Vamos derivar a função!
![f(x)=2 x^{2} -5\\\\\
f'(x)=4x\\\\\
Sendo~~ x_{0}=2~~e~~so~~substituir~~na~~func\~ao.\\\\\\
f'(2)=4(2)\\\\\
f'(2)=8\\\\\\
\boxed{Resposta:f'(2)=8}](https://tex.z-dn.net/?f=f%28x%29%3D2+x%5E%7B2%7D+-5%5C%5C%5C%5C%5C%0Af%27%28x%29%3D4x%5C%5C%5C%5C%5C%0ASendo%7E%7E+x_%7B0%7D%3D2%7E%7Ee%7E%7Eso%7E%7Esubstituir%7E%7Ena%7E%7Efunc%5C%7Eao.%5C%5C%5C%5C%5C%5C%0Af%27%282%29%3D4%282%29%5C%5C%5C%5C%5C%0Af%27%282%29%3D8%5C%5C%5C%5C%5C%5C%0A%0A%5Cboxed%7BResposta%3Af%27%282%29%3D8%7D%0A%0A+ "f(x)=2 x^{2} -5\\\\\
f'(x)=4x\\\\\
Sendo~~ x_{0}=2~~e~~so~~substituir~~na~~func\~ao.\\\\\\
f'(2)=4(2)\\\\\
f'(2)=8\\\\\\
\boxed{Resposta:f'(2)=8}")
É só isso!
Solução!
Vamos derivar a função!
É só isso!
marypereira1210:
show!
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