# Consider the function 𝑓(𝑥)=5−2𝑥^2, −3≤𝑥≤1.?

The absolute maximum value is

and this occurs at 𝑥 equals

The absolute minimum value is

and this occurs at 𝑥 equals

Relevance

f(x) = -2x² + 5

Find the absolute min and max on the interval of [-3, 1].

Let's start with finding the values of f(x) at these endpoints to get a baseline:

f(-3) = -2(-3)² + 5 and f(1) = -2(1)² + 5

f(-3) = -2(9) + 5 and f(1) = -2(1) + 5

f(-3) = -18 + 5 and f(1) = -2 + 5

f(-3) = -13 and f(1) = 3

So far we have a max at f(1) = 3 and a min at f(-3) = -13.

Next, we solve for any zeroes of the first derivative and check any results in the range above.  Then check those values of f(x):

f(x) = -2x² + 5

f'(x) = -4x

0 = -4x

0 = x

There is one root.  Now we see what f(0) is and compare it to our current min and max values:

f(x) = -2x² + 5

f(0) = -2(0)² + 5

f(0) = -2(0) + 5

f(0) = 0 + 5

f(0) = 5

That's a new max.  So the absolute extrema values are:

min at f(-3) = -13

max at f(0) = 5

• This is a 2nd deg poly, a "parabola". So, w/o calculus, you should be able to answer your questions. Use this as a way to verify your answers via calclulus.

Via calculus:

Absolute max & mins are found by verifying the points f '(x) = 0 AND boundary points.

So, what is your f '(x) ?

Solving for f '(x) = 0 gives ...?