Calculus Question

Question

Find the absolute maximum and absolute minimum values of f on the given interval.

f(x) = x⁴ − 8x² + 6,    [−3, 4]

absolute minimum:

absolute maximum:

Answer 1

f(x) = x⁴ − 8x² + 6

f'(x) = 3x^3 - 16x = 0

x = 0; 3x^2-16 = 0; x^2 = 16/3, x = +/- 4/sqrt(3)

0 is the only zero within the [-3,4] range. Check if max or min by plugging test points into derivative on either side of point.

f'(-1) = -3+16 = positive, so increasing

f'(1) = 3-16 = negative, so decreasing

At zero, function goes from increasing to decreasing. Therefore, this is an absolute max. To find absolute min, check the other two points.

f(x) = x⁴ − 8x² + 6

f(-3) = 81-72+6 = 15 - this is lower, therefore, absolute min

f(4) = 256-128+6 = 134

If this doesn't make sense, let me know and we can jump on a quick call.

Answer 2

To find the absolute minimum and/or maximum on an interval, we need to do two things.

1) Find any local extremes that are within the interval. One of those may be the absolute minimum and/or maximum.

2) Check at the boundaries of the interval. One of those could be the absolute minimum and/or maximum.

First, to find any local max/min on the interval, we take the first derivative and set it equal to zero.

f '(x) = 4x³- 16x

4x³- 16x = 0 [factor out 4x to get:

4x(x² - 4) = 0 [factor the difference of two squares to get:

4x(x - 2)(x + 2) = 0 [set each piece equal to zero to get the three "zeros":

x = 0, x = 2, x = -2. All three of these are on the interval in question.

So now, set up a table with the local max/mins as well as the endpoints (our second criteria) and calculate the function values at each point like this:

x | f(x)

-3 | 15

-2 | -10

0 | 6

2 | -10

4 | 134

So, we can see from the table that the absolute minimum occurs at both x = -2 and x = 2 and is y = -10 and the absolute maximum occurs at x = 4 and is y = 134.

So we say that the absolute minimum value is -10 and the absolute maximum value is 134 on the interval [−3, 4]

Answer 3

The determine mins and maxes, take the derivative and find the critical values, where the derivative or slope =0

f'(x) = 4x³ − 16x Factor

=4x(x²-4)

=4x(x-2)(x+2) Setting each factor equal to zero and solving

Critical values are x=-2,0,2

Now you need to pick test points along each of these intervals, to see if the derivative is positive or negative. You create a sing chart to do this. When the slope changes from positive to negative, that is a relative max, when it changes from negative to positive that is a relative min.

I have 3 factors. Each factor I plug in the test point and see what the sign is for each factor. then multiply them together to determine the sign of the derivative.

Test point -3 = (-)(-)(-) = negative (3 negatives multiplied together is negative)

Test point -1 = (-)(-)(+) = positive

Test point 1 = (+)(-)(+) = negative

Test point 3 = (+)(+)(+) = positive

What this tells is:

At the first CV, the slope goes from negative to positive, x=-2 is a relative min

At the second CV, the slope goes from positive to negative, x=0 is a relative max

At the third CV, the slope goes from negative to positive, x=2 is a relative min

In addition to these points, we also have to look at the endpoints -3 and 4. Here is a summary of all the possible mins and maxes

(-3,15)

(-2,-10)

(0,6)

(2,-10)

(4,134)

The absolute max value is 134 and it occurs at (4,134)

The absolute min value of the function is -10, it occurs at 2 points in the domain (-2,-10) and (2,-10)