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

Sep 16, 2015

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

#### Explanation:

$f \left(x\right) = {x}^{2} - 2 x - 3$, 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 $\left(- \infty , 1\right) \mathmr{and} \left(1 , + \infty\right)$

Now consider any value of x in $\left(- \infty , 1\right)$, 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