Question

Find the absolute maximum value of p(x)=−x2−x 1 over [−1,1].

Answer 1

The absolute maximum value of
`p(x)` over the interval [-1, 1] is 1, and it occurs at x = -1/2.

The critical point you found is x = -1/2. To find the absolute maximum or minimum on the interval [-1, 1], we also need to evaluate the function at the endpoints of the interval.

Let's find p(-1):

`p(-1)=-(-1)^2-(-1)+1=-1+1+1=1`

Now, let's find p(1):

`p(1)=-(1)^2-(1)+1=-1-1+1=-1`

So, we have values at the critical point x = -1/2, x = -1, and x = 1:

`\begin{aligned}& p(-1 / 2)=-(-1 / 2)^2-(-1 / 2)+1 \\& p(-1)=1 \\& p(1)=-1\end{aligned}`

Comparing these values, we can see that the maximum value occurs at x= -1/2, where
`p(-1 / 2)=1`.

Therefore, the absolute maximum value of
`p(x)=-x^2-x+1` over the interval [-1, 1] is 1, and it occurs at x = -1/2.

Answer 2

The absolute maximum in the given interval is (-0.5, 0.25)

How to find the absolute maximum value?

We have a quadratic function:

`p(x) = -x^2 - x`

And we want to find the maximum. Notice that this is a quadratic equation which opens downwards, so the maximum is at the vertex of the parabola.

The x-value of the vertex is:

x = -(-1)/2*(-1) = -0.5

Evaluating there we get:

`p(-0.5) = -(-0.5)^2 + 0.5 = -0.25 + 0.5 = 0.25`

And x = -0.5 is in the given interval, so the maximum is (-0.5, 0.25)