How do use the first derivative test to determine the local extrema #x^2-2x-3#?

1 Answer
Sep 16, 2015

Answer:

There is a minima of f(x) at x= 1

Explanation:

#f(x)= x^2 -2x-3#, hence f'(x)= 2x-2.

Since f(x) is defined for all values of x, the only critical point is given by 2x-2=0, or, x=1

Since divide the entire domain in two parts #(-oo,1) and (1,+oo)#

Now consider any value of x in #(-oo,1)#, say, x=-1. For x= -1, f'(x) would equal to -2-2= -4. This means the slope of the curve is negative.

At x=-1, the slope is 0

At any point in #(1, +oo), say x=2, f'(x) would be 4-2=2. This means the slope of the curve is positive.

The above analysis of the slope of f(x) , shows that the slope of the curve changes from negative, to the left of x=1, to positive to its right. This implies that there is a minima at x=1