Question

How do you find the exact relative maximum and minimum of the polynomial function of `y=x^2 +x -1`?

Answer 1

we have a minimum at `(-1/2,-5/4)`

Explanation:

To find a min/max we look for values of `x` that make the derivative vanish. To determine the nature of those turning points we perform the second derivative test

We are dealing with a positive quadratic so we expect a single minimum.

We have:

`y = x^2 + x -1`

Differentiating wrt `x` we get:

`dy/dx = 2x+1`

For the derivative to vanish we have:

`dy/dx = 0 => 2x+1 =0`
`:. x=-1/2`

Differentiating again wrt `x` we get:

`(d^2y)/(dx)^2 = 2`

So when `x=-1/2 => (d^2y)/(dx)^2 > 0` confirming a minimum.

Finally, When `x=-1/2 => y= 1/4-1/2-1 = -5/4`

Thus we have a minimum at `(-1/2,-5/4)`

graph{y = x^2 + x -1 [-10, 10, -5, 5]}