Monday

October 20, 2014

October 20, 2014

Posted by **Robert** on Monday, October 8, 2007 at 12:43am.

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**, Monday, October 8, 2007 at 2:28am55. 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**, Tuesday, October 9, 2007 at 1:00am67. dy/dx = 0

Would it be all the points of the function.

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