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3 years ago

If f(x) = x2 - 2x and g(x) = 6x + 4 for which value of x does (t +g)(x)=0

Mathematics

1 answer:

Vinil7 [7]3 years ago

7 0

Answer:

-2

Assumption:

Find the value of x such that $(f+g)(x)=0$ .

Step-by-step explanation:

$(f+g)(x)=0$

$f(x)+g(x)=0$

$(x^2-2x)+(6x+4)=0$

Combine like terms:

$x^2+4x+4=0$

This is not too bad too factor on the left hand side since 2(2)=4 and 2+2=4.

$(x+2)(x+2)=0$

$(x+2)^2=0$

So we need to solve:

$x+2=0$

Subtract 2 on both sides:

$x=-2$

Let's check:

$(f+g)(-2)$

$f(-2)+g(-2)$

$((-2)^2-2(-2))+(6(-2)+4)$

$(4+4)+(-12+4)$

$(8)+(-8)$

$0$

0 was the desired output of $(f+g)(x)$ .

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3 years ago

(x^2y+e^x)dx-x^2dy=0

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It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$