# Find the absolute maximum value and the absolute minimum value for the following function:...

## Question:

Find the absolute maximum value and the absolute minimum value for the following function:

{eq}f(x)=-x^2-x+6 \ \text {on} \ [-1,3] {/eq}

## Absolute maximum and minimum values of a function

This example illustrates how to find the absolute maximum and minimum values of a function over a given interval.

It is important to note that differentiation yields only local maxima or minima, and these values can sometimes be bettered.

At a local maximum or minimum, {eq}f'(x) = 0. {/eq}

This gives:

{eq}-2x - 1 = 0 \\ \Rightarrow x = -\dfrac{1}{2} \\ \Rightarrow f(-\dfrac{1}{2}) = -(-\dfrac{1}{2})^2 - (-\dfrac{1}{2}) + 6 = \dfrac{25}{4} {/eq}

Finding the second derivative:

{eq}f''(x) = -2. {/eq}

Since this is negative, {eq}(-\dfrac{1}{2}, \dfrac{25}{4}) {/eq} is a local maximum.

We must now also check the end points of the given domain {eq}-1 \leq x \leq 3. {/eq}

{eq}f(-1) = -(-1)^2 - (-1) + 6 = 6 \\ f(3) = -(3)^2 - (3) + 6 = -6 {/eq}

Looking at the function values at the ends of the interval, together with the local maximum, we see that

{eq}-6 < 6 < \dfrac{25}{4}. {/eq}

So the absolute minimum value is -6,

and the absolute maximum value is {eq}\dfrac{25}{4}. {/eq}