use the first dervative test to determine the local extrema.

Question

Use the first derivative test to determine the local extrema If the function f, then verify you answer with a graph.

1. F(x)= x^2(x-1)(x+1)

Answer 1

  

The function is F(x) = x²(x - 1)(x + 1).

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

Apply Sum and Difference of Same Terms : (u + v)(u - v) = u² - v².

F(x) = x²(x² - 1²)

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

F(x) = x⁴ - x²

Apply first derivative with respect to x.

F'(x) = 4x³ - 2x.

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

f^,(x) = 0

4x³ - 2x = 0

2x(2x² - 1) = 0

2x = 0 and 2x² - 1 = 0 ⇒ x = ± √(1/2) = ± √(0.5) = ± 0.707.

The critical numbers are x = 0 and ± √(1/2), The test intervals are (-∞, -0.707), (-0.707, 0), (0, 0.707) and (0.707, +∞)

(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
(-∞, -0.707)	x = -1	F'(-1) = 4(-1)3 - 2(-1) = -2 < 0	Decreasing
(-0.707, 0)	x = -0.1	F'(-0.1) = 4(-0.1)3 - 2(-0.1) = 0.196 > 0	Increasing
(0, 0.707)	x = 0.1	F'(0.1) = 4(0.1)3 - 2(0.1) = -0.196 < 0	Decreasing
(0.707, +∞)	x = 0.1	F'(1) = 4(1)3 - 2(1) = 2> 0	Increasing

(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 minimum at x = -√(0.5), because f'(x) changes from negative to positive around -√(0.5).

F[-√(0.5)] = (-√0.5)²[(-√0.5)² - 1] = 0.5[0.5 - 1] = -0.25.

The relative minimum point (-√0.5, -0.25).

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

F(0) = (0)²[0² - 1] = 0.

The relative maximum point (0, 0).

The function f(x) has relative minimum at x = √(0.5), because f'(x) changes from negative to positive around √(0.5).

F[√(0.5)] = (√0.5)²[(√0.5)² - 1] = 0.5[0.5 - 1] = - 0.25.

The relative minimum point (√0.5, - 0.25).

Graph :