jme12 wrote:Given the function: f(x)=x^{4}-8^{2}+7

1. At what point(s), if any, are the tangents to the graph of f(x) horizontal?

So I found the derivative of f(x) and set it to zero...

0=4x^{3}-16x

0=4x(x^{2}-4)

0=4x(x+2)(x-2)

and then set the three parts of the equation equal to zero to find the x values..

0=4x --> x=0

0=x+2 --> x=-2

0=x-2 --> x=2

and now do I plug these values into the original equation to find y to give me the points?

Exactly: Now that you've found the x-values for which f'(x) is zero (and thus the tangents are horizontal), you need to plug those x-values back into the original function to find the corresponding y-values, and thus the points on the graph.

jme12 wrote:2. What is f'(x)>0? When is f'(x)<0? What does the value of f'(x) tell you about the graph o f(x)?

You have the

**polynomial inequality** 4x(x + 2)(x - 2) > 0. You know that the graph of f'(x) is going to be a cubic, coming up from the left, dipping around in the middle, and then heading up on the right.

Where will this graph cross the x-axis, splitting the number line into intervals? What are these intervals? Where is the graph above the x-axis? So on which intervals is f'(x) positive?

Answer the same sorts of questions to find where f'(x) < 0.

For what this says about f(x), you'll need to study up on

**derivative and slope** (that is, how the sign of the derivative relates to how the function is increasing or decreasing).

jme12 wrote:3. Determine the point(s) on f(x) where f"(x)=0?

The second derivative is just the first derivative of the first derivative, so do the same thing you did for the first part (above), but start with f'(x) instead of f(x).

Eliz.