Question on the slope of the tangent

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Find the points on the graph of y= (1/3)x^3-5x- (4/x) at which the slope of the tangent is horizontal.

what i know:
- we have to use m=[f(a+h)-f(a)]/h
- if we change the equation we can get 3x^4 - 15x^2 -12
- the slope of the tangent is zero.

THANX
 
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hint: how does the derivative of a function relate to the slope of the tangent of its graph?
 
Find the derivative, and set it equal to the slope of a horizontal line. What is that?
 
What everyone is saying is "DO IT"! By the way are you really required to use the "quotient difference" formula? It's not too difficult but tedious and most problems like this allow the use of derivative formulas.
 
Gib Z said:
Find the derivative, and set it equal to the slope of a horizontal line. What is that?

Do you mean, "what is the slope of a horizontal line?"

Draw a graph that shows a horizontal line. Pick two points (x_1,y_1) and (x_2,y_2) on the line. Do you know how to calculate the slope of a line from two points?
 
yes that's what i meant, but i knew the answer..set the derivative to zero is what i meant
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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