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muminat
3 years ago
12

Determine whether each relation is a function

Mathematics
1 answer:
Alex3 years ago
7 0

Step-by-step explanation:

(2,4),(3,5),(4,6),(5,8) (1,4),(2,3),(3,-6),(-3,6)

is NOT a function, because some x values (2 and 3) have more than one associated result value (4, 3 and 5, -6).

(1,5),(1,7),(1,-5),(1,-7) (0,4),(1,4),(2,4),(3,4)

is NOT a function as the x value 1 has more than 1 associated result values (5, 7, -5, -7, 4).

x² = y + 1

y = x² - 1

this is a function. for any value of x we get exactly one result value.

xy = 1

y = 1/x

this is a function. for any value of x we get exactly one result value. but we do need to exclude x=0.

if we cannot do that, then it is NOT a function, because x=0 has no associated result value.

x = 3

is NOT a function, because although there is only one valid x value (3), there are infinitely many possible y result values associated with this x value (y can be "anything").

y = -2

this is a function. for any value of x we get exactly one result value (constantly -2).

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Presumably you should be doing this using calculus methods, namely computing the surface integral along \mathbf r(u,v).

But since \mathbf r(u,v) describes a sphere, we can simply recall that the surface area of a sphere of radius a is 4\pi a^2.

In calculus terms, we would first find an expression for the surface element, which is given by

\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv

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=4\pi a^2

as expected.
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