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 

2 Answers

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  • 4 weeks ago
    Favorite Answer

    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

  • rotchm
    Lv 7
    4 weeks ago

    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 ...?

    Your boundary points are?

    Evaluating f(x) at all those pints, which one is the max and min?

    Done!

    Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them. 

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