Find the relative extreme values of the function f(x) = x 3 3 − x 2 2 − 6x + 5.

Answer 1

I'm going to assume the function is:

f(x) = x³ - x² - 6x + 5

The derivative finds the slope of the line tangent to the function. If the line tangent is horizontal (a slope of zero), then the function must have a minimum or maximum at that point.

So, take the derivative:

f '(x) = 3x² - 2x - 6

This provides the slope of the tangent line at any value of "x". To find out the values of "x" where the slopes are zero, set the derivative function equal to zero and solve for the associated x-values:

3x² - 2x - 6 = 0

Using the quadratic formula:

x = [-(-2) ± √((-2)² - 4(3)(-6))]/(2•3)

x = [2 ± √76]/6 = (2 ± 2√19)/6 = (1 ± √19)/3

So the possible extremes occur at x = (1 - √19)/3 and at x = (1 + √19)/3. To find out for sure, we can check the slope of the tangent line on each side of these "critical points"

(1 - √19)/3 is approx - 1.1 so let's calculate the slope at x = -2:

3(-2)² - 2(-2) - 6 = 12 + 4 - 6 = 10 so the slope is positive on this interval.

At x = 0 the slope is:

3(0)² - 2(0) - 6 = -6 so the slope is negative on this interval. That means since the slope changes from positive to negative that x = (1 - √19)/3 is a local maximum.

Since (1 + √19)/3 is approx 1.8, lets check the slope at x = 2:

3(2)² - 2(2) - 6 = 12 - 4 - 6 = 2 so the slope is positive on this interval. That means x = (1 + √19)/3 must be a local minimum since the slope changes from negative to positive.

Note that it is also possible that the slope doesn't change signs at a critical point meaning that although the slope is zero at the critical point, it is not a local extreme. That's why we need to check.

Now, finally, to find the extreme values, you must plug in x = (1 - √19)/3 and x = (1 + √19)/3 into the original function f(x) = x³ - x² - 6x + 5 to get the function values. I'll let you do that.