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3 years ago

If f(x) = (x + 7)^2, which of the following is g(x) based on the translation?

Mathematics

2 answers:

Nady [450]3 years ago

5 0

Answer:

The answer is the 2nd option :)

Step-by-step explanation:

The only thing I can tell you is that my answer is the right one ;)

lana66690 [7]3 years ago

3 0

F(x) = (x+7)²
(x+7)(x+7) = x(x+7)+7(x+7) = x² + 7x + 7x + 49 = x² + 14x + 49

(x+9)² = (x+9)(x+9) = x(x+9)+9(x+9) = x² + 9x + 9x + 81 = x² + 18x + 81
(x+5)² = (x+5)(x+5) = x(x+5)+5(x+5) = x² + 5x + 5x + 25 = x² + 10x + 25
(x-9)² = (x-9)(x-9) = x(x-9)-9(x-9) = x² - 9x - 9x + 81 = x² - 18x + 81
(x-5)² = (x-5)(x-5) = x(x-5)-5(x-5) = x² - 5x - 5x + 25 = x² - 10x + 25

I think g(x) based on the translation would be g(x) = (x+9)²

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To take advantage of the characteristic solutions $y_1(x)=x$ and $y_2(x)=e^x$ , you can try the method of variation of parameters, where we look for a solution of the form

$y=y_1u_1+y_2u_2$

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Then

$y'={y_1}'u_1+y_1{u_1}'+{y_2}'u_2+y_2{u_2}'$

$\implies y'={y_1}'u_1+{y_2}'u_2$

$y''={y_1}''u_1+{y_1}'{u_1}'+{y_2}''u_2+{y_2}'{u_2}'$

Substituting into the ODE gives

$(1-x)({y_1}''u_1+{y_1}'{u_1}'+{y_2}''u_2+{y_2}'{u_2}')+x({y_1}'u_1+{y_2}'u_2)-y_1u_1+y_2u_2=2(x-1)^2e^{-x}$

Since

$y_1=x\implies{y_1}'=1\implies{y_1}''=0$

$y_2=e^x\implies{y_2}'=e^x\implies{y_2}''=e^x$

the above reduces to

$(1-x)({u_1}'+e^x{u_2}')=2(x-1)^2e^{-x}$

${u_1}'+e^x{u_2}'=2(1-x)e^{-x}$ $(\mathbf 2)$

$(\mathbf 1)$ and $(\mathbf 2)$ form a linear system that we can solve for ${u_1}',{u_2}'$ using Cramer's rule:

${u_1}'=\dfrac{W_1(x)}{W(x)},{u_2}'=\dfrac{W_2(x)}{W(x)}$

where $W(x)$ is the Wronskian determinant of the fundamental system and $W_i(x)$ is the same determinant, but with the $i$ -th column replaced with $(0,2(x-1)^2e^{-x})$ .