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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? 

 Aug 27, 2020
 #1
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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.

 Aug 27, 2020
 #2
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Hi thank you for the reply. It took me a bit to understand how to do it but I think I got it! So does this mean that the minimum is always zero, seeing as multiplying by zero will always result in that same number?

Guest Aug 27, 2020
 #3
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It isn't a minimum of the original cubic function, it is the point of steepest negative slope.  It is where the changing slope has a minimum. Try plotting the function for a range of x values, and then plot the slope over the same range of x values ( say, from x = -2 to 2)  to see.

Alan  Aug 27, 2020

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