Alan

I need to calculate the point in which the graph of f(x) = (1/3)x^3 - 4x + 2 declines the most using the derivative. Could anyone help me? 

The derivative is df(x) = x^2 -4

To find the minimum of df(x) we differentiate again to find ddf(x) = 2x

To find where this is a minimum we set it equal to zero.  Clearly it is zero when x = 0, so this is the point at which f(x) has the steepest negative slope.

Hint:  4 is also a common factor of the two terms within parentheses in the denominator.

Like so:

As follows:

First think of the inner f(x) as, say, y.

f(y) = 5    

so either y² - 4 = 5 or y + 3 = 5  

If y² - 4 = 5, then y = 3 is a possibility as 3 > -4, also y = -3 is a possibility as -3 > -4

If y + 3 = 5 then  y = 2, which is not possible as 2 is not less than -4

So there are two possibilities for y, which means you have to consider two possibilities for f(x)

(1)  f(x) = 3

      3 = x² - 4  means x² = 7  or x = ±√7  both of which are > -4

       3 = x + 3  means x = 0  which is not possible because 0 is not less than -4.

(2) f(x) = -3

     Can you do this one?

Yes, I gave the equation of the line.  gwen... has correctly supplied the inequality.

The intercept is -4 and the slope is 4/2, so, in slope intercept form the equation is y = 2x - 4

I'll leave you to rearrange this in "standard" form (a pointless exercise in my view!).  

If x = 1, then f(1) = 0 is real

If x = 2 then the denominator is 0, so f(2) is undefined.

I wonder if you meant to say the smallest positive (rather than possible) integer, in which case

if x = 3 then f(3) = sqrt(2) which is real.

The following should give you enough information to do the calculations:

This graph should help (if you can see it!)