Download Problem 1: First derivative: Productrule

Problem 1: First derivative: Productrule
Problem 2: First derivative and Domain of Definition
Problem 3: First derivative
Problem 4: First derivative
Problem 5: Local Maxima/Minima
T(x) = -x² + 4x + 4
T(x) = 
1 2
x x4
4
T(x) = -2,25x² + 4,5x +18
T(x) =
1 2
x  2x  3
2
T(x) = 2x²-6x + 65
-(x - 2)² + 8
TMax = 8 für x = 2
1
 ( x  2)²  5
4
TMax = 5 für x = 2
-2,25 (x -1)² + 20,25
TMax = 20,25 für x = 1
1
( x  2)²  1
2
TMin = 1 für x = 2
2(x-1,5)² + 60,5
TMin = 60,5 für x = 1,5
T(x) =
2 2
x 3
5
2
( x  0) 2  3
5
TMin = -3 für x = 0
T(x) =
1 2
x  2x  1
2
1
( x  2)²  1
2
TMin = -1 für x = -2
T(x) = 1,5 x2 - 5
1,5 (x - 0)2 - 5
TMin = -5 für x = 0
T(x) = -0,25x² + 3x -5,25
-0,25(x - 6)² + 3,75
TMax = 3,75 für x = 6
4
 ( x  2,5)²  120,33
3
TMax =120,33 für x= 2,5
-(x -3)² + 7
TMax = 7 für x = 3
T(x) = 
4 2 20
x 
x  112
3
3
T(x) = -x² + 6x - 2
Problem 6: Tangent Line
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.
Problem 2: Find a and b so that the line y = -3x + 4 is tangent to the graph of y = ax3
+ bx 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
Problem 7: Tangent Line
1 - Find all points on the graph of y = x 3 - 3x where the tangent line is parallel to the
line whose equation is given by y = 9x + 4.
2 - Find a and b so that the line y = -2 is tangent to the graph of y = ax2 + bx at x = 1.
3 - Find conditions on a, b and c so that the graph of y = ax 3 + bx 2 + cx has ONE
tangent line parallel to the x axis (horizontal tangent).
solutions to the above exercises
1-
(2,2) and (-2,-2)
2-
a = 2 and b = - 4
3-
4b 2 - 12 ac = 0
Problem 8: Limits (Grenzwerte)
Find the limits of f(x), where x-> -∞ and x-> ∞
Problem 9: Limits (Grenzwerte)