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3 years ago

For which nonnegative value of x is the expression 5+x

Mathematics

1 answer:

Nikitich [7]3 years ago

6 0

Answer:

x = 5

Step-by-step explanation:

the expression is "undefined" when the denominator is equal to zero

$\frac{\left(5+x\right)}{25-x^2\:} \\$

denominator = 25 - x²

Values of x where the equation is "undefined"

0 = 25 - x²

x² = 25

√x² = √25

x = ± 5

Nonnegative value of x where the equation is "undefined"

x = 5

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

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 :

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