How do use the first derivative test to determine the local extrema #(x^2-10x)^4#?

1 Answer
Aug 28, 2015

Answer:

The function has local (and global) minima at #x=0# and #x=10# and a local maximum at #x=5#.

Explanation:

Let #f(x)=(x^2-10x)^4#, then the Power Rule and Chain Rule give a derivative of

#f'(x)=4(x^2-10x)^3*(2x-10)#

#=8x^3(x-10)^3(x-5)#

#f'(x)# is therefore clearly zero at #x=0,5,10#, and these are the critical points of #f#.

A bit less clearly, #f'# changes sign from negative to positive as #x# increases through #x=0#, from positive to negative as #x# increases through #x=5#, and from negative to positive as #x# increases through #x=10#. You can plug numbers less than 0, between 0 and 5, between 5 and 10, and greater than 10 and use the continuity of #f'# to check this. You can also graph #f'# to confirm it graphically.

The First Derivative Test then implies that the function has local (and global) minima at #x=0# and #x=10# and a local maximum at #x=5#.

Interestingly, application of the Second Derivative Test in this example only allows us to conclude where the local maximum is, not where the local minima are (it turns out that #f''(0)=f''(10)=0# and #f''(5)=-125000<0#). In other words, the First Derivative Test is "stronger" (it can be used in more situations) than the Second Derivative Test.