How do use the first derivative test to determine the local extrema #x^2-x-1#?

1 Answer
Sep 12, 2015

Answer:

#(1/2,-5/4)#.

Explanation:

The "peaks" or local extrema of a function #f(x)# occur at the values where #d/dx f(x) = 0#.

One way to remember this is:
Since a function is increasing when #d/dx f(x) > 0#, and a function is decreasing when #d/dx f(x) < 0#, that means when #d/dx f(x) = 0#, the graph of the function is "turning" from increasing to decreasing or from decreasing to increasing. The turn forms a "peak" which is a local extrema.

So, to get the values for the local extrema for #f(x) = x^2 - x - 1#, we need to evaluate #f(x)# at the values of #x# where #d/dx f(x) = 0#.

First, we get the derivative using the power rule:
#color(green)(d/dx x^n = nx^(n-1))#

#f(x) = x^2 - x - 1#

#d/dx f(x) = 2x - 1#

Then, we solve for the values where #d/dx f(x) = 0#:

#0 = 2x - 1#
#1 = 2x#
#1/2 = x#

So, there is a local extrema when #x = 1/2#.

To find the value of the local extrema, we evaluate #f(1/2)#.

#f(x) = x^2 - x - 1#
#f(1/2) = (1/2)^2 - 1/2 - 1#
#f(1/2) = 1/4 - 1/2 - 1#
#f(1/2) = -5/4#

There exists a local extrema at the point #(1/2,-5/4)#.