Find the exact location of all the relative and absolute extrema of the function. f(x) = -x³ - 6x² - 9x - 2.

Solution:

Given f(x) = -x³ - 6x² - 9x - 2.

For a given function, relative extrema, or local maxima and minima, can be determined by using the first derivative test, which allows you to check for any sign changes of f′ around the function's critical points.

f(x) = -x³ - 6x² - 9x - 2

For a critical point to be local extrema, the function must go from increasing, i.e. positive f′, and to decrease, i.e. negative f′, or vice versa, around that point.

Let’s determine the first derivative of f.

f ′= -3x² - 12x - 9

To determine the function's critical points, make f′=0 and solve for x

f′ = -3x² - 12x - 9 = 0

This is equivalent to -3(x² + 4x + 3) = 0, or x² + 4x + 3 = 0

x = {-4 ± √4² - 4⋅1⋅3} / 2

x = {-4 ± √16 - 12 } / 2

x = {-4 ± √4 } /2

x = {-4 ± 2 } /2

x = -3, -1

Since no domain restrictions are given for the function, both solutions will be critical points.

Now let’s check to see if the first derivative changes sign around these points.

Since we have got two critical points, we will evaluate them in 3 intervals.

Select a value from each of these intervals and note the sign of f′

For (-∞, -3):

f′(-4) = -3⋅(-4 + 1)⋅(-4 + 3)

f′(-4) = -3⋅(-3)⋅(-1) = -9 → negative

For (-3, -1):

f′(0) = -3⋅(-2 + 1)⋅(-2 + 3)

f′(0) = -3⋅(-1)⋅(+1) = 3 → positive

For (-1, ∞):

f′(0) = -3⋅(0 + 1)⋅(0 + 3)

f′(0) = -3⋅1⋅3 = -9 → negative

→ The first derivative changes sign twice. It goes from being negative to being positive around x = -3, which means that this critical point is a local minimum.

→ On the other hand, it goes from being positive to being negative around point x = -1, which means that this critical point is a local maximum.

→ This is equivalent to having a function that goes from decreasing to increasing (think of a valley) around point x = -3, and from increasing to decreasing (think of a hill) around point x = -1

→ To get the actual points at which the function has the local minimum and maximum, evaluate f at the critical points.

f(-3) = -(-3)³ - 6(-3)² - 9(-3) - 2

f(3) = 27 - 54 - 27 - 2 = -2 and

f(-1) = -(-1)³ - 6(-1)² - 9(-1) - 2

f(-1) = 1 - 6 + 9 - 2 = 2

Therefore, the function f has(-3, -2) → local minimum(-1, 2) → local maximum

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Find the exact location of all the relative and absolute extrema of the function. f(x) = -x³ - 6x² - 9x - 2

Summary:

The exact location of all the relative and absolute extrema of the function.f(x) = -x³ - 6x² - 9x - 2 are (-3,-2) and (-1,2).