Posted by Robert on .
55. Find an equation of the tangent line to the graph of
f(x) =x^3  2x + 1
a) at (2,5);
b) at (1,2);
c) at (0,1).
I believe the answers are:
a) y= (3/2 x)  (3/2)
b) No solution
c) No solution
For each function, find the points on the graph at which the tangent line is horizontal. If none exists, state that fact.
67. y=4
73. f(x) = (1/3)^x^3 + (1/2)^x^2  2
For the function, find the points on the graph at which the tangent line has slope 1.
79. y=(1/3)^x^3 + (2)^x^2 + 2x

Calculus 
drwls,
55. The slope any point of the line is the derivative,
df/dx = 3x^2 2
When x = 2, f(x) = 5 and
df/dx = (3*4)  2 = 10
The coordinates of (b) and (c) are also on the line, and there IS a slope there. Use the same df/dx formuyla to compute it.
I will be happy to provide assistance with you other three problems when work is shown. In #79, are you sure that x^3 is a power of 2 and x^2 is a power of 1/2? It looks to me like you used too many ^ signs. 
Calculus 
Robert,
67. dy/dx = 0
Would it be all the points of the function.