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تاريخ التسجيل : 21/09/2008
 |  موضوع: مسائل tangent lines in culculas  2008-09-26, 8:50 pm | |
| | <table><tr vAlign=top><td vAlign=top>Problem 1: Find all points on the graph of y = x 3 - 3x where the tangent line is parallel to the x axis (or horizontal tangent line).
Solution to Problem 1:
- Lines that are parallel to the x axis have slope = 0. The slope of a tangent line to the graph of y = x 3 - 3x is given by the first derivative y '.
y ' = 3x 2 - 3
- We now find all values of x for which y ' = 0.
3x 2 - 3 = 0
- Solve the above equation for x to obtain the solutions.
x = -1 and x = 1
- The above values of x are the x coordinates of the points where the tangent lines are parallel to the x axis. Find the y coordinates of these points using y = x 3 - 3x
for x = -1 , y = 2
for x = 1 , y = -2
- The points at which the tangent lines are parallel to the x axis are: (-1,2) and (1,-2). See the graph of y = x3 - 3x below with the tangent lines.
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Problem 2: Find a and b so that the graph of y = ax3 + bx is tangent to the line y = -3x + 4 at x = 1.
Solution to Problem 2:
- We need to determine two algebraic equations in order to find a and b. Since the point of tangency is on the graph of y = ax3 + bx and y = -3x + 4, at x = 1 we have
a(1)3 + b(1) = -3(1) + 4
- Simplify to write an equation in a and b
a + b = 1
- The slope of the tangent line is -3 which is also equal to the first derivative y ' of y = ax3 + bx at x = 1
y ' = 3ax2 + x = -3 at x = 1.
- The above gives a second equation in a and b
3a + b = -3
- Solve the system of equations a + b = 1 and 3a + b = -3 to find a and b
a = -2 and b = 3.
- See graphs of y = ax3 + bx, with a = -2 and b = 3, and y = -3x + 4 below.
Problem 3: Find conditions on a and b so that the graph of y = ae x + bx has NO tangent line parallel to the x axis (horizontal tangent).
Solution to Problem 3:
- The slope of a tangent line is given by the first derivative y ' of y = ae x + bx. Find y '
y ' = ae x + b
- To find the x coordinate of a point at which the tangent line to the graph of y is horizontal, solve y ' = 0 for x (slope of a horizontal line = 0)
ae x + b = 0
- Rewrite the above equation as follows
e x = -b/a
- The above equation has solutions for -a/b >0. Hence, the graph of y = ae x + bx has NO horizontal tangent line if -a/b <= 0
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