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3 years ago

How do you do this question?

Mathematics

1 answer:

Kitty [74]3 years ago

4 0

Answer:

C) One vertical but no horizontal

Step-by-step explanation:

Use implicit differentiation to find dy/dx.

xy² = 2 + xy

x (2y dy/dx) + y² = x dy/dx + y

(2xy − x) dy/dx = y − y²

dy/dx = (y − y²) / (2xy − x)

When the tangent lines are horizontal, dy/dx = 0.

0 = (y − y²) / (2xy − x)

0 = y − y²

0 = y (1 − y)

y = 0 or 1

When the tangent lines are vertical, dy/dx is undefined, so the denominator is 0.

0 = 2xy − x

0 = x (2y − 1)

x = 0, y = 1/2

There are two possible points where there can be a horizontal tangent line, and two possible points where there can be a vertical tangent line. Plug all back into the original equation to see which are actual points.

If y = 0:

x(0)² = 2 + x(0)

0 = 2

No solution.

If y = 1:

x(1)² = 2 + x(1)

x = 2 + x

0 = 2

No solution.

If x = 0:

(0)y² = 2 + (0)y

0 = 2

No solution.

If y = 1/2:

x(1/2)² = 2 + x(1/2)

x/4 = 2 + x/2

-x/4 = 2

x = -8

Therefore, there is one vertical line and no horizontal lines.

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Explanation:

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2 years ago

Find a function where f(0)=2 and f(1)=2

xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

$f(x) = 2$ is a valid solution.

Want something more fun? Why not a parabola? $f(x)= ax^2+bx+c$ .

At this point you have three parameters to play with, and from the fact that $f(0)=2$ we can already fix one of them, in particular $c=2$ . At this point I would recommend picking an easy value for one of the two, let's say $a= 1$ (or even $a=-1$ , it will just flip everything upside down) and find out b accordingly: $f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1$

Our function becomes

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Notice that it works even by switching sign in the first two terms: $f(x) = -x^2+x+2$

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: $f(x) = A cos (kx)$

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need $A= 2$ , and at that point the first condition is guaranteed; using the second to find k we get $2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi$