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Find the x-value at which the tangent to the function of y= (4/e^2x-2) + 8x is parallel to the x-axis.

User ThrawnCA
by
6.5k points

1 Answer

4 votes

Answer:

Explanation:

The tangent line is parallel to the

x

axis when the slope (hence

d

y

d

x

) is zero and it is parallel to the

y

axis when the slope (again,

d

y

d

x

) goes to

or

Explanation:

We'll start by finding

d

y

d

x

:

x

2

+

x

y

+

y

2

=

7

d

d

x

(

x

2

+

x

y

+

y

2

)

=

d

d

x

(

7

)

2

x

+

1

y

+

x

d

y

d

x

+

2

y

d

y

d

x

=

0

d

y

d

x

=

2

x

+

y

x

+

2

y

Now,

d

y

d

x

=

0

when the nuimerator is

0

, provided that this does not also make the denominator

0

.

2

x

+

y

=

0

when

y

=

2

x

We have now, two equations:

x

2

+

x

y

+

y

2

=

7

y

=

2

x

Solve (by substitution)

x

2

+

x

(

2

x

)

+

(

2

x

)

2

=

7

x

2

2

x

2

+

4

x

2

=

7

3

x

2

=

7

x

=

±

7

3

=

±

21

3

Using

y

=

2

x

, we get

The tangent to the curve is horizontal at the two points:

(

21

3

,

2

21

3

)

and

(

21

3

,

2

21

3

)

(Observe that these pair do not also make the denominator of

d

y

d

x

equal to

0

)

To find the points at which the tangent is vertical, make the denominator of

d

y

d

x

equal tpo

0

(without also making the numerator

0

).

We could go through the solution, but the symmetry of the equation that we will get:

x

=

2

y

, so

y

=

±

21

3

and the points on the curve at which the tangent is vertical are:

(

2

21

3

,

21

3

)

and

(

2

21

3

,

21

3

)

By the way. Because we do have the technology, here is the graph of this rotated ellipse: (Note that

±

21

3

±

1.528

which you can see on the graph.)

graph{x^2 + xy +y^2 =7 [-11.3, 11.2, -5.665, 5.585]}

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User Cero
by
6.0k points