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

1 Answer
Steve M
Jun 26, 2017

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]}

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