Vertical tangent lines If a function f is continuous at a and lim x → a | f ′ ( x ) | = ∞ , then the curse y = f ( x ) has a vertical tangent line at a and the equation of the tangent line is x = a. If a is an endpoint of a domain, then the appropriate one-sided derivative (Exercises 31–32) is used. Use this information to answer the following questions. 36. Graph the following curves and determine the location of any vertical tangent lines. a. x 2 + y 2 = 9 b. x 2 + y 2 + 2 x = 0
Vertical tangent lines If a function f is continuous at a and lim x → a | f ′ ( x ) | = ∞ , then the curse y = f ( x ) has a vertical tangent line at a and the equation of the tangent line is x = a. If a is an endpoint of a domain, then the appropriate one-sided derivative (Exercises 31–32) is used. Use this information to answer the following questions. 36. Graph the following curves and determine the location of any vertical tangent lines. a. x 2 + y 2 = 9 b. x 2 + y 2 + 2 x = 0
Solution Summary: The author illustrates how the function x2+y 2=9 has a vertical tangent at x=3, and the derivative is infinite at these points.
Vertical tangent linesIf a function f is continuous at a and
lim
x
→
a
|
f
′
(
x
)
|
=
∞
, then the curse y = f(x) has a vertical tangent line at a and the equation of the tangent line is x = a. If a is an endpoint of a domain, then the appropriate one-sided derivative (Exercises 31–32) is used. Use this information to answer the following questions.
36. Graph the following curves and determine the location of any vertical tangent lines.
Find a plane containing the point (3, -3, 1) and the line of intersection of the planes 2x + 3y - 3z = 14
and -3x - y + z = −21.
The equation of the plane is:
Determine whether the lines
L₁ : F(t) = (−2, 3, −1)t + (0,2,-3) and
L2 : ƒ(s) = (2, −3, 1)s + (−10, 17, -8)
intersect. If they do, find the point of intersection.
● They intersect at the point
They are skew lines
They are parallel or equal
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Differential Equation | MIT 18.01SC Single Variable Calculus, Fall 2010; Author: MIT OpenCourseWare;https://www.youtube.com/watch?v=HaOHUfymsuk;License: Standard YouTube License, CC-BY