To find the relative extrema of the function, we need to find the critical points by taking the derivative of the function and setting it equal to zero.
f(x) = x^3 + 2x^2 - 2x
f'(x) = 3x^2 + 4x - 2
Setting f'(x) = 0:
3x^2 + 4x - 2 = 0
To solve for x, we can use the quadratic formula:
x = (-4 ± √(4^2 - 4(3)(-2))) / (2(3))
x = (-4 ± √(16 + 24)) / 6
x = (-4 ± √40) / 6
x = (-4 ± 2√10) / 6
x = (-2 ± √10) / 3
So the critical points are (-2 + √10) / 3 and (-2 - √10) / 3.
To determine if each critical point is a relative maximum or minimum, we can use the second derivative test. Taking the second derivative:
f''(x) = 6x + 4
Evaluating f'' at each critical point:
f''((-2 + √10) / 3) = 6(-2 + √10) / 3 + 4 = -2 + 2√10 + 4 = 2√10 + 2
f''((-2 - √10) / 3) = 6(-2 - √10) / 3 + 4 = -2 - 2√10 + 4 = -2√10 + 2
Since f''((-2 + √10) / 3) is positive, the point (-2 + √10) / 3 is a relative minimum.
Since f''((-2 - √10) / 3) is negative, the point (-2 - √10) / 3 is a relative maximum.
Now we can evaluate the function at each critical point to find the coordinates of the relative extrema:
f((-2 + √10) / 3) = ((-2 + √10) / 3)^3 + 2((-2 + √10) / 3)^2 - 2(-2 + √10) / 3
f((-2 + √10) / 3) ≈ 5.064
f((-2 - √10) / 3) = ((-2 - √10) / 3)^3 + 2((-2 - √10) / 3)^2 - 2(-2 - √10) / 3
f((-2 - √10) / 3) ≈ -1.647
Rounding to the nearest hundredth, we have the relative maximum at (-2.68, -1.65) and the relative minimum at (-0.75, 5.06). Therefore, the correct answer is:
(-2.68, -1.65)
(-0.75, 5.06)
Find the relative maximum and relative minimum of the graph of the function, f(x) = x^3 + 2x^2 - 2x. Determine where the relative maximum occurs and round each coordinate to the nearest hundredth. Show your work.
(0.39,-0.42)
(-2.68,-1.65)
(-0.75,5.06)
(-1.72,4.27)
1 answer