Question

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

Answer 1

`(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)`.