consider the function f(x) = (x+2)^2 (x-1)

Question

a) find the critical numbers of fx

b)find the open intervals on which the function is increasing or decreasing

c) apply the first derivative test to identify all relative extrema

Answer 1

(a).

The function is f(x) = (x + 2)²(x - 1).

Write the above equation in general form of polynomial function P(x) = a_nxⁿ + a_n-1x^n-1 + . . . + a₁x + a₀.

Apply Square of a Binomial : (u + v)² = u² + 2uv + v².

f(x) = [x² + 2(2)(x) + 2²](x - 1)

f(x) = (x² + 4x + 4)(x - 1)

f(x) = (x - 1)(x² + 4x + 4)

Apply Distributive property : a(b - c) = ab - ac.

f(x) = x(x² + 4x + 4) - (x² + 4x + 4)

f(x) = x³ + 4x² + 4x - x² - 4x - 4

f(x) = x³ + 3x² - 4

To find the critical or key numbers, to make the first derivative equal to zero [ f ' (x) = 0] or f ' (x) does not exist.

Apply first derivative with respect to x.

f^,(x) = 3x² + 6x.

f^,(x) = 0

3x² + 6x = 0

3x(x + 2) = 0

The critical numbers are x = 0 and x = - 2.

(b).

TEST FOR INCREASING AND DECREASING FUNCTIONS :

If f^,(x) > 0 (Positive) for all x in (a, b), then f(x) is increasing on [a, b].

If f^,(x) < 0 (Negative) for all x in (a, b), then f(x) is decreasing on [a, b].

Test intervals    x - Value                    Sign of f^,(x)                             Conclusion

(- ∞, - 2)              x = - 3         3(- 3)² + 6(- 3) = 27 - 18 = 9 > 0        Increasing

(- 2, 0)                 x = - 1         3(- 1)² + 6(- 1) = 3 - 6 = - 3 < 0          Decreasing

(0, ∞)                   x = 1           3(1)² + 6(1) = 3 + 6 = 9 > 0                Increasing

So, f(x) is increasing on the interval (- ∞, - 2) and (0, ∞) and decreasing on the interval (- 2, 0).

(c).

Find Relative Extrema :

The First Derivative Test :

Let f be a differential function with f(c) = 0 then

• If f ' (x) changes from positive to negative, f has a relative (local) maximum at c.
• If f ' (x) changes from negative to positive, f has a relative (local) minimum at c.

The function f(x) has relative maximum at x = - 2, because f'(x) changes from positive to negative around − 2.

f( - 2) = (- 2)³ + 3(- 2)² - 4 = - 8 + 12 - 4 = 0.

The relative maximum point (- 2, 0).

The function f(x) has relative minimum at x = 0, because f'(x) changes from negative to positive around 0.

f(0) = (0)³ + 3(0)² - 4 = 0 + 0 - 4 = - 4.

The relative minimum point (0, - 4).

Graph :